CALCULUS I - Google Sites · PDF fileCalculus I © 2007 Paul Dawkins ii Preface Here are...

CALCULUS I Derivatives

Paul Dawkins

Calculus I

© 2007 Paul Dawkins i http://tutorial.math.lamar.edu/terms.aspx

Table of Contents Preface ............................................................................................................................................. ii Derivatives ...................................................................................................................................... 3

Introduction ................................................................................................................................................ 3 The Definition of the Derivative ................................................................................................................ 5 Interpretations of the Derivative............................................................................................................... 11 Differentiation Formulas .......................................................................................................................... 20 Product and Quotient Rule ....................................................................................................................... 28 Derivatives of Trig Functions................................................................................................................... 34 Derivatives of Exponential and Logarithm Functions .............................................................................. 45 Derivatives of Inverse Trig Functions ...................................................................................................... 50 Derivatives of Hyperbolic Functions ....................................................................................................... 56 Chain Rule ................................................................................................................................................ 58 Implicit Differentiation ............................................................................................................................ 68 Related Rates ............................................................................................................................................ 77 Higher Order Derivatives ......................................................................................................................... 91 Logarithmic Differentiation ..................................................................................................................... 96

http://tutorial.math.lamar.edu/terms.aspx

Calculus I

© 2007 Paul Dawkins ii http://tutorial.math.lamar.edu/terms.aspx

Preface Here are my online notes for my Calculus I course that I teach here at Lamar University. Despite the fact that these are my “class notes”, they should be accessible to anyone wanting to learn Calculus I or needing a refresher in some of the early topics in calculus. I’ve tried to make these notes as self contained as possible and so all the information needed to read through them is either from an Algebra or Trig class or contained in other sections of the notes. Here are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed.

1. Because I wanted to make this a fairly complete set of notes for anyone wanting to learn calculus I have included some material that I do not usually have time to cover in class and because this changes from semester to semester it is not noted here. You will need to find one of your fellow class mates to see if there is something in these notes that wasn’t covered in class.

2. Because I want these notes to provide some more examples for you to read through, I don’t always work the same problems in class as those given in the notes. Likewise, even if I do work some of the problems in here I may work fewer problems in class than are presented here.

3. Sometimes questions in class will lead down paths that are not covered here. I try to anticipate as many of the questions as possible when writing these up, but the reality is that I can’t anticipate all the questions. Sometimes a very good question gets asked in class that leads to insights that I’ve not included here. You should always talk to someone who was in class on the day you missed and compare these notes to their notes and see what the differences are.

4. This is somewhat related to the previous three items, but is important enough to merit its own item. THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!! Using these notes as a substitute for class is liable to get you in trouble. As already noted not everything in these notes is covered in class and often material or insights not in these notes is covered in class.


Calculus I

© 2007 Paul Dawkins 3 http://tutorial.math.lamar.edu/terms.aspx

Derivatives

Introduction In this chapter we will start looking at the next major topic in a calculus class. We will be looking at derivatives in this chapter (as well as the next chapter). This chapter is devoted almost exclusively to finding derivatives. We will be looking at one application of them in this chapter. We will be leaving most of the applications of derivatives to the next chapter. Here is a listing of the topics covered in this chapter. The Definition of the Derivative – In this section we will be looking at the definition of the derivative. Interpretation of the Derivative – Here we will take a quick look at some interpretations of the derivative. Differentiation Formulas – Here we will start introducing some of the differentiation formulas used in a calculus course. Product and Quotient Rule – In this section we will look at differentiating products and quotients of functions. Derivatives of Trig Functions – We’ll give the derivatives of the trig functions in this section. Derivatives of Exponential and Logarithm Functions – In this section we will get the derivatives of the exponential and logarithm functions. Derivatives of Inverse Trig Functions – Here we will look at the derivatives of inverse trig functions. Derivatives of Hyperbolic Functions – Here we will look at the derivatives of hyperbolic functions. Chain Rule – The Chain Rule is one of the more important differentiation rules and will allow us to differentiate a wider variety of functions. In this section we will take a look at it. Implicit Differentiation – In this section we will be looking at implicit differentiation. Without this we won’t be able to work some of the applications of derivatives.


Calculus I


Related Rates – In this section we will look at the lone application to derivatives in this chapter. This topic is here rather than the next chapter because it will help to cement in our minds one of the more important concepts about derivatives and because it requires implicit differentiation. Higher Order Derivatives – Here we will introduce the idea of higher order derivatives. Logarithmic Differentiation – The topic of logarithmic differentiation is not always presented in a standard calculus course. It is presented here for those who are interested in seeing how it is done and the types of functions on which it can be used.


Calculus I


The Definition of the Derivative In the first section of the last chapter we saw that the computation of the slope of a tangent line, the instantaneous rate of change of a function, and the instantaneous velocity of an object at x a= all required us to compute the following limit.

( ) ( )limx a

f x f ax a→

−−

We also saw that with a small change of notation this limit could also be written as,

( ) ( )0

limh

f a h f ah→

+ − (1)

This is such an important limit and it arises in so many places that we give it a name. We call it a derivative. Here is the official definition of the derivative. Definition The derivative of ( )f x with respect to x is the function ( )f x′ and is defined as,

( ) ( ) ( )0

limh

f x h f xf x

h→

+ −′ = (2)

Note that we replaced all the a’s in (1) with x’s to acknowledge the fact that the derivative is really a function as well. We often “read” ( )f x′ as “f prime of x”.

Let’s compute a couple of derivatives using the definition. Example 1 Find the derivative of the following function using the definition of the derivative. ( ) 22 16 35f x x x= − + Solution So, all we really need to do is to plug this function into the definition of the derivative, (1), and do some algebra. While, admittedly, the algebra will get somewhat unpleasant at times, but it’s just algebra so don’t get excited about the fact that we’re now computing derivatives. First plug the function into the definition of the derivative.

( ) ( ) ( )

( ) ( ) ( )0

2 2

0

lim

2 16 35 2 16 35lim

h

h

f x h f xf x

hx h x h x x

h

→

→

+ −′ =

+ − + + − − +=

Be careful and make sure that you properly deal with parenthesis when doing the subtracting. Now, we know from the previous chapter that we can’t just plug in 0h = since this will give us a


Calculus I


division by zero error. So we are going to have to do some work. In this case that means multiplying everything out and distributing the minus sign through on the second term. Doing this gives,

( )

2 2 2

0

2

0

2 4 2 16 16 35 2 16 35lim

4 2 16lim

h

h

x xh h x h x xf xh

xh h hh

→

→

+ + − − + − + −′ =

+ −=

Notice that every term in the numerator that didn’t have an h in it canceled out and we can now factor an h out of the numerator which will cancel against the h in the denominator. After that we can compute the limit.

( ) ( )0

0

4 2 16lim

lim 4 2 16

4 16

h

h

h x hf x

hx h

x

→

→

+ −′ =

= + −

= −

So, the derivative is, ( ) 4 16f x x′ = − Example 2 Find the derivative of the following function using the definition of the derivative.

( )1

tg tt

=+

Solution This one is going to be a little messier as far as the algebra goes. However, outside of that it will work in exactly the same manner as the previous examples. First, we plug the function into the definition of the derivative,

( ) ( ) ( )

0

0

lim

1lim1 1

h

h

g t h g tg t

ht h t

h t h t

→

→

+ −′ =

+ = − + + +

Note that we changed all the letters in the definition to match up with the given function. Also note that we wrote the fraction a much more compact manner to help us with the work. As with the first problem we can’t just plug in 0h = . So we will need to simplify things a little. In this case we will need to combine the two terms in the numerator into a single rational expression as follows.


Calculus I


( ) ( )( ) ( )( ) ( )

( )( ) ( )

( ) ( )

0

2 2

0

0

1 11lim1 1

1lim1 1

1lim1 1

h

h

h

t h t t t hg t

h t h t

t t th h t th th t h t

hh t h t

→

→

→

+ + − + +′ = + + +

+ + + − + + = + + +

= + + +

Before finishing this let’s note a couple of things. First, we didn’t multiply out the denominator. Multiplying out the denominator will just overly complicate things so let’s keep it simple. Next, as with the first example, after the simplification we only have terms with h’s in them left in the numerator and so we can now cancel an h out. So, upon canceling the h we can evaluate the limit and get the derivative.

( ) ( ) ( )

( )( )

( )

0

2

1lim1 1

11 1

11

hg t

t h t

t t

t

→′ =

+ + +

=+ +

=+

The derivative is then,

( )( )2

11

g tt

′ =+

Example 3 Find the derivative of the following function using the definition of the derivative. ( ) 5 8R z z= − Solution First plug into the definition of the derivative as we’ve done with the previous two examples.

( ) ( ) ( )

( )0

0

lim

5 8 5 8lim

h

h

R z h R zR z

hz h z

h

→

→

+ −′ =

+ − − −=

In this problem we’re going to have to rationalize the numerator. You do remember rationalization from an Algebra class right? In an Algebra class you probably only rationalized the denominator, but you can also rationalize numerators. Remember that in rationalizing the numerator (in this case) we multiply both the numerator and denominator by the numerator except we change the sign between the two terms. Here’s the rationalizing work for this problem,


Calculus I


( )( )( ) ( )( )

( )( )( )

( )( )

( )( )

0

0

0

5 8 5 8 5 8 5 8lim

5 8 5 8

5 5 8 5 8lim

5 8 5 8

5lim5 8 5 8

h

h

h

z h z z h zR z

h z h z

z h z

h z h z

hh z h z

→

→

→

+ − − − + − + −′ =

+ − + −

+ − − −=

+ − + −

=+ − + −

Again, after the simplification we have only h’s left in the numerator. So, cancel the h and evaluate the limit.

( )( )0

5lim5 8 5 8

55 8 5 8

52 5 8

hR z

z h z

z z

z

→′ =

+ − + −

=− + −

=−

And so we get a derivative of,

( ) 52 5 8

R zz

′ =−

Let’s work one more example. This one will be a little different, but it’s got a point that needs to be made. Example 4 Determine ( )0f ′ for ( )f x x= Solution Since this problem is asking for the derivative at a specific point we’ll go ahead and use that in our work. It will make our life easier and that’s always a good thing. So, plug into the definition and simplify.

( ) ( ) ( )0

0

0

0 00 lim

0 0lim

lim

h

h

h

f h ff

hhh

hh

→

→

→

+ −′ =

+ −=

=


Calculus I


We saw a situation like this back when we were looking at limits at infinity. As in that section we can’t just cancel the h’s. We will have to look at the two one sided limits and recall that

if 0if 0

h hh

h h≥

= − <

( )0 0

0

lim lim because 0 in a left-hand limit.

lim 1

1

h h

h

h h hh h− −

−

→ →

→

−= <

= −

= −

0 0

0

lim lim because 0 in a right-hand limit.

lim 1

1

h h

h

h h hh h+ +

+

→ →

→

= >

=

=

The two one-sided limits are different and so

0

limh

hh→

doesn’t exist. However, this is the limit that gives us the derivative that we’re after. If the limit doesn’t exist then the derivative doesn’t exist either. In this example we have finally seen a function for which the derivative doesn’t exist at a point. This is a fact of life that we’ve got to be aware of. Derivatives will not always exist. Note as well that this doesn’t say anything about whether or not the derivative exists anywhere else. In fact, the derivative of the absolute value function exists at every point except the one we just looked at, 0x = . The preceding discussion leads to the following definition. Definition A function ( )f x is called differentiable at x a= if ( )f x′ exists and ( )f x is called

differentiable on an interval if the derivative exists for each point in that interval. The next theorem shows us a very nice relationship between functions that are continuous and those that are differentiable. Theorem If ( )f x is differentiable at x a= then ( )f x is continuous at x a= .


Calculus I


See the Proof of Various Derivative Formulas section of the Extras chapter to see the proof of this theorem. Note that this theorem does not work in reverse. Consider ( )f x x= and take a look at,

( ) ( )

0 0lim lim 0 0x x

f x x f→ →

= = =

So, ( )f x x= is continuous at 0x = but we’ve just shown above in Example 4 that

( )f x x= is not differentiable at 0x = .

Alternate Notation Next we need to discuss some alternate notation for the derivative. The typical derivative notation is the “prime” notation. However, there is another notation that is used on occasion so let’s cover that. Given a function ( )y f x= all of the following are equivalent and represent the derivative of

( )f x with respect to x.

( ) ( )( ) ( )df dy d df x y f x ydx dx dx dx

′ ′= = = = =

Because we also need to evaluate derivatives on occasion we also need a notation for evaluating derivatives when using the fractional notation. So if we want to evaluate the derivative at x=a all of the following are equivalent.

( ) x ax a x a

df dyf a ydx dx=

= =

′ ′= = =

Note as well that on occasion we will drop the (x) part on the function to simplify the notation somewhat. In these cases the following are equivalent. ( )f x f′ ′= As a final note in this section we’ll acknowledge that computing most derivatives directly from the definition is a fairly complex (and sometimes painful) process filled with opportunities to make mistakes. In a couple of sections we’ll start developing formulas and/or properties that will help us to take the derivative of many of the common functions so we won’t need to resort to the definition of the derivative too often. This does not mean however that it isn’t important to know the definition of the derivative! It is an important definition that we should always know and keep in the back of our minds. It is just something that we’re not going to be working with all that much.


Calculus I


Interpretations of the Derivative Before moving on to the section where we learn how to compute derivatives by avoiding the limits we were evaluating in the previous section we need to take a quick look at some of the interpretations of the derivative. All of these interpretations arise from recalling how our definition of the derivative came about. The definition came about by noticing that all the problems that we worked in the first section in the chapter on limits required us to evaluate the same limit. Rate of Change The first interpretation of a derivative is rate of change. This was not the first problem that we looked at in the limit chapter, but it is the most important interpretation of the derivative. If

( )f x represents a quantity at any x then the derivative ( )f a′ represents the instantaneous rate

of change of ( )f x at x a= .

Example 1 Suppose that the amount of water in a holding tank at t minutes is given by

( ) 22 16 35V t t t= − + . Determine each of the following. (a) Is the volume of water in the tank increasing or decreasing at 1t = minute?

[Solution]

(b) Is the volume of water in the tank increasing or decreasing at 5t = minutes? [Solution]

(c) Is the volume of water in the tank changing faster at 1t = or 5t = minutes? [Solution]

(d) Is the volume of water in the tank ever not changing? If so, when? [Solution] Solution In the solution to this example we will use both notations for the derivative just to get you familiar with the different notations. We are going to need the rate of change of the volume to answer these questions. This means that we will need the derivative of this function since that will give us a formula for the rate of change at any time t. Now, notice that the function giving the volume of water in the tank is the same function that we saw in Example 1 in the last section except the letters have changed. The change in letters between the function in this example versus the function in the example from the last section won’t affect the work and so we can just use the answer from that example with an appropriate change in letters. The derivative is.

( ) 4 16 OR 4 16dVV t t tdt

′ = − = −


Calculus I


Recall from our work in the first limits section that we determined that if the rate of change was positive then the quantity was increasing and if the rate of change was negative then the quantity was decreasing. We can now work the problem. (a) Is the volume of water in the tank increasing or decreasing at 1t = minute? In this case all that we need is the rate of change of the volume at 1t = or,

( )1

1 12 OR 12t

dVVdt =

′ = − = −

So, at 1t = the rate of change is negative and so the volume must be decreasing at this time.

[Return to Problems] (b) Is the volume of water in the tank increasing or decreasing at 5t = minutes? Again, we will need the rate of change at 5t = .

( )5

5 4 OR 4t

dVVdt =

′ = =

In this case the rate of change is positive and so the volume must be increasing at 5t = .

[Return to Problems] (c) Is the volume of water in the tank changing faster at 1t = or 5t = minutes? To answer this question all that we look at is the size of the rate of change and we don’t worry about the sign of the rate of change. All that we need to know here is that the larger the number the faster the rate of change. So, in this case the volume is changing faster at 1t = than at 5t = .

[Return to Problems] (d) Is the volume of water in the tank ever not changing? If so, when? The volume will not be changing if it has a rate of change of zero. In order to have a rate of change of zero this means that the derivative must be zero. So, to answer this question we will then need to solve

( ) 0 OR 0dVV tdt

′ = =

This is easy enough to do. 4 16 0 4t t− = ⇒ = So at 4t = the volume isn’t changing. Note that all this is saying is that for a brief instant the volume isn’t changing. It doesn’t say that at this point the volume will quit changing permanently.


Calculus I


If we go back to our answers from parts (a) and (b) we can get an idea about what is going on. At 1t = the volume is decreasing and at 5t = the volume is increasing. So at some point in time

the volume needs to switch from decreasing to increasing. That time is 4t = . This is the time in which the volume goes from decreasing to increasing and so for the briefest instant in time the volume will quit changing as it changes from decreasing to increasing.

[Return to Problems] Note that one of the more common mistakes that students make in these kinds of problems is to try and determine increasing/decreasing from the function values rather than the derivatives. In this case if we took the function values at 0t = , 1t = and 5t = we would get, ( ) ( ) ( )0 35 1 21 5 5V V V= = = Clearly as we go from 0t = to 1t = the volume has decreased. This might lead us to decide that AT 1t = the volume is decreasing. However, we just can’t say that. All we can say is that between 0t = and 1t = the volume has decreased at some point in time. The only way to know what is happening right at 1t = is to compute ( )1V ′ and look at its sign to determine

increasing/decreasing. In this case ( )1V ′ is negative and so the volume really is decreasing at

1t = . Now, if we’d plugged into the function rather than the derivative we would have gotten the correct answer for 1t = even though our reasoning would have been wrong. It’s important to not let this give you the idea that this will always be the case. It just happened to work out in the case of 1t = . To see that this won’t always work let’s now look at 5t = . If we plug 1t = and 5t = into the volume we can see that again as we go from 1t = to 5t = the volume has decreased. Again, however all this says is that the volume HAS decreased somewhere between 1t = and 5t = . It does NOT say that the volume is decreasing at 5t = . The only way to know what is going on right at 5t = is to compute ( )5V ′ and in this case ( )5V ′ is positive and so the volume is

actually increasing at 5t = . So, be careful. When asked to determine if a function is increasing or decreasing at a point make sure and look at the derivative. It is the only sure way to get the correct answer. We are not looking to determine is the function has increased/decreased by the time we reach a particular point. We are looking to determine if the function is increasing/decreasing at that point in question.


Calculus I


Slope of Tangent Line This is the next major interpretation of the derivative. The slope of the tangent line to ( )f x at

x a= is ( )f a′ . The tangent line then is given by,

( ) ( )( )y f a f a x a′= + − Example 2 Find the tangent line to the following function at 3z = . ( ) 5 8R z z= − Solution We first need the derivative of the function and we found that in Example 3 in the last section. The derivative is,

( ) 52 5 8

R zz

′ =−

Now all that we need is the function value and derivative (for the slope) at 3z = .

( ) ( ) 53 7 32 7

R m R′= = =

The tangent line is then,

( )57 32 7

y z= + −

Velocity Recall that this can be thought of as a special case of the rate of change interpretation. If the position of an object is given by ( )f t after t units of time the velocity of the object at t a= is

given by ( )f a′ .

Example 3 Suppose that the position of an object after t hours is given by,

( )1

tg tt

=+

Answer both of the following about this object. (a) Is the object moving to the right or the left at 10t = hours? [Solution] (b) Does the object ever stop moving? [Solution]

Solution Once again we need the derivative and we found that in Example 2 in the last section. The derivative is,

( )( )2

11

g tt

′ =+

(a) Is the object moving to the right or the left at 10t = hours? To determine if the object is moving to the right (velocity is positive) or left (velocity is


Calculus I


negative) we need the derivative at 10t = .

( ) 110121

g′ =

So the velocity at 10t = is positive and so the object is moving to the right at 10t = .

[Return to Problems] (b) Does the object ever stop moving? The object will stop moving if the velocity is ever zero. However, note that the only way a rational expression will ever be zero is if the numerator is zero. Since the numerator of the derivative (and hence the speed) is a constant it can’t be zero. Therefore, the velocity will never stop moving. In fact, we can say a little more here. The object will always be moving to the right since the velocity is always positive.

[Return to Problems] We’ve seen three major interpretations of the derivative here. You will need to remember these, especially the rate of change, as they will show up continually throughout this course. Before we leave this section let’s work one more example that encompasses some of the ideas discussed here and is just a nice example to work. Example 4 Below is the sketch of a function ( )f x . Sketch the graph of the derivative of this

function, ( )f x′ .

Solution At first glance this seems to an all but impossible task. However, if you have some basic knowledge of the interpretations of the derivative you can get a sketch of the derivative. It will not be a perfect sketch for the most part, but you should be able to get most of the basic features


Calculus I


of the derivative in the sketch. Let’s start off with the following sketch of the function with a couple of additions.

Notice that at 3x = − , 1x = − , 2x = and 4x = the tangent line to the function is horizontal. This means that the slope of the tangent line must be zero. Now, we know that the slope of the tangent line at a particular point is also the value of the derivative of the function at that point. Therefore, we now know that, ( ) ( ) ( ) ( )3 0 1 0 2 0 4 0f f f f′ ′ ′ ′− = − = = =

This is a good starting point for us. It gives us a few points on the graph of the derivative. It also breaks the domain of the function up into regions where the function is increasing and decreasing. We know, from our discussions above, that if the function is increasing at a point then the derivative must be positive at that point. Likewise, we know that if the function is decreasing at a point then the derivative must be negative at that point. We can now give the following information about the derivative.

( )( )( )( )( )

3 0

3 1 0

1 2 0

2 4 0

4 0

x f x

x f x

x f x

x f x

x f x

′< − <

′− < < − >

′− < < <

′< < <

′> >

Remember that we are giving the signs of the derivatives here and these are solely a function of whether the function is increasing or decreasing. The sign of the function itself is completely immaterial here and will not in any way effect the sign of the derivative. This may still seem like we don’t have enough information to get a sketch, but we can get a little


Calculus I


bit more information about the derivative from the graph of the function. In the range 3x < − we know that the derivative must be negative, however we can also see that the derivative needs to be increasing in this range. It is negative here until we reach 3x = − and at this point the derivative must be zero. The only way for the derivative to be negative to the left of 3x = − and zero at 3x = − is for the derivative to increase as we increase x towards 3x = − . Now, in the range 3 1x− < < − we know that the derivative must be zero at the endpoints and positive in between the two endpoints. Directly to the right of 3x = − the derivative must also be increasing (because it starts at zero and then goes positive – therefore it must be increasing). So, the derivative in this range must start out increasing and must eventually get back to zero at

1x = − . So, at some point in this interval the derivative must start decreasing before it reaches 1x = − . Now, we have to be careful here because this is just general behavior here at the two

endpoints. We won’t know where the derivative goes from increasing to decreasing and it may well change between increasing and decreasing several times before we reach 1x = − . All we can really say is that immediately to the right of 3x = − the derivative will be increasing and immediately to the left of 1x = − the derivative will be decreasing. Next, for the ranges 1 2x− < < and 2 4x< < we know the derivative will be zero at the endpoints and negative in between. Also, following the type of reasoning given above we can see in each of these ranges that the derivative will be decreasing just to the right of the left hand endpoint and increasing just to the left of the right hand endpoint. Finally, in the last region 4x > we know that the derivative is zero at 4x = and positive to the right of 4x = . Once again, following the reasoning above, the derivative must also be increasing in this range. Putting all of this material together (and always taking the simplest choices for increasing and/or decreasing information) gives us the following sketch for the derivative.

Note that this was done with the actual derivative and so is in fact accurate. Any sketch you do


Calculus I


will probably not look quite the same. The “humps” in each of the regions may be at different places and/or different heights for example. Also, note that we left off the vertical scale because given the information that we’ve got at this point there was no real way to know this information. That doesn’t mean however that we can’t get some ideas of specific points on the derivative other than where we know the derivative to be zero. To see this let’s check out the following graph of the function (not the derivative, but the function).

At 2x = − and 3x = we’ve sketched in a couple of tangent lines. We can use the basic rise/run slope concept to estimate the value of the derivative at these points. Let’s start at 3x = . We’ve got two points on the line here. We can see that each seem to be about one-quarter of the way off the grid line. So, taking that into account and the fact that we go through one complete grid we can see that the slope of the tangent line, and hence the derivative, is approximately -1.5. At 2x = − it looks like (with some heavy estimation) that the second point is about 6.5 grids above the first point and so the slope of the tangent line here, and hence the derivative, is approximately 6.5. Here is the sketch of the derivative with the vertical scale included and from this we can see that in fact our estimates are pretty close to reality.


Calculus I


Note that this idea of estimating values of derivatives can be a tricky process and does require a fair amount of (possible bad) approximations so while it can be used, you need to be careful with it. We’ll close out this section by noting that while we’re not going to include an example here we could also use the graph of the derivative to give us a sketch of the function itself. In fact, in the next chapter where we discuss some applications of the derivative we will be looking using information the derivative gives us to sketch the graph of a function.


Calculus I


Differentiation Formulas In the first section of this chapter we saw the definition of the derivative and we computed a couple of derivatives using the definition. As we saw in those examples there was a fair amount of work involved in computing the limits and the functions that we worked with were not terribly complicated. For more complex functions using the definition of the derivative would be an almost impossible task. Luckily for us we won’t have to use the definition terribly often. We will have to use it on occasion, however we have a large collection of formulas and properties that we can use to simplify our life considerably and will allow us to avoid using the definition whenever possible. We will introduce most of these formulas over the course of the next several sections. We will start in this section with some of the basic properties and formulas. We will give the properties and formulas in this section in both “prime” notation and “fraction” notation. Properties

1) ( ) ( )( ) ( ) ( )f x g x f x g x′ ′ ′± = ± OR ( ) ( )( )d df dgf x g xdx dx dx

± = ±

In other words, to differentiate a sum or difference all we need to do is differentiate the individual terms and then put them back together with the appropriate signs. Note as well that this property is not limited to two functions. See the Proof of Various Derivative Formulas section of the Extras chapter to see the proof of this property. It’s a very simple proof using the definition of the derivative.

2) ( )( ) ( )cf x cf x′ ′= OR ( )( )d dfcf x cdx dx

= , c is any number

In other words, we can “factor” a multiplicative constant out of a derivative if we need to. See the Proof of Various Derivative Formulas section of the Extras chapter to see the proof of this property.

Note that we have not included formulas for the derivative of products or quotients of two functions here. The derivative of a product or quotient of two functions is not the product or quotient of the derivatives of the individual pieces. We will take a look at these in the next section. Next, let’s take a quick look at a couple of basic “computation” formulas that will allow us to actually compute some derivatives.


Calculus I


Formulas

1) If ( )f x c= then ( ) 0f x′ = OR ( ) 0d cdx

=

The derivative of a constant is zero. See the Proof of Various Derivative Formulas section of the Extras chapter to see the proof of this formula.

2) If ( ) nf x x= then ( ) 1nf x nx −′ = OR ( ) 1n nd x nxdx

−= , n is any number.

This formula is sometimes called the power rule. All we are doing here is bringing the original exponent down in front and multiplying and then subtracting one from the original exponent. Note as well that in order to use this formula n must be a number, it can’t be a variable. Also note that the base, the x, must be a variable, it can’t be a number. It will be tempting in some later sections to misuse the Power Rule when we run in some functions where the exponent isn’t a number and/or the base isn’t a variable. See the Proof of Various Derivative Formulas section of the Extras chapter to see the proof of this formula. There are actually three different proofs in this section. The first two restrict the formula to n being an integer because at this point that is all that we can do at this point. The third proof is for the general rule, but does suppose that you’ve read most of this chapter.

These are the only properties and formulas that we’ll give in this section. Let’s compute some derivatives using these properties. Example 1 Differentiate each of the following functions.

(a) ( ) 100 1215 3 5 46f x x x x= − + − [Solution]

(b) ( ) 6 62 7g t t t−= + [Solution]

(c) 35

18 233

y z zz

= − + − [Solution]

(d) ( ) 3 7

5 2

29T x x xx

= + − [Solution]

(e) ( ) 2h x x xπ= − [Solution] Solution (a) ( ) 100 1215 3 5 46f x x x x= − + −

In this case we have the sum and difference of four terms and so we will differentiate each of the terms using the first property from above and then put them back together with the proper sign. Also, for each term with a multiplicative constant remember that all we need to do is “factor” the constant out (using the second property) and then do the derivative.


Calculus I


( ) ( ) ( ) ( )99 11 0

99 11

15 100 3 12 5 1 0

1500 36 5

f x x x x

x x

′ = − + −

= − +

Notice that in the third term the exponent was a one and so upon subtracting 1 from the original exponent we get a new exponent of zero. Now recall that 0 1x = . Don’t forget to do any basic arithmetic that needs to be done such as any multiplication and/or division in the coefficients.

[Return to Problems] (b) ( ) 6 62 7g t t t−= +

The point of this problem is to make sure that you deal with negative exponents correctly. Here is the derivative.

( ) ( ) ( )5 7

5 7

2 6 7 6

12 42

g t t t

t t

−

−

′ = + −

= −

Make sure that you correctly deal with the exponents in these cases, especially the negative exponents. It is an easy mistake to “go the other way” when subtracting one off from a negative exponent and get 56t−− instead of the correct 76t−− .

[Return to Problems]

(c) 35

18 233

y z zz

= − + −

Now in this function the second term is not correctly set up for us to use the power rule. The power rule requires that the term be a variable to a power only and the term must be in the numerator. So, prior to differentiating we first need to rewrite the second term into a form that we can deal with.

3 518 233

y z z z−= − + −

Note that we left the 3 in the denominator and only moved the variable up to the numerator. Remember that the only thing that gets an exponent is the term that is immediately to the left of the exponent. If we’d wanted the three to come up as well we’d have written,

( )5

13z

so be careful with this! It’s a very common mistake to bring the 3 up into the numerator as well at this stage. Now that we’ve gotten the function rewritten into a proper form that allows us to use the Power Rule we can differentiate the function. Here is the derivative for this part.

2 6524 13

y z z−′ = + +



Calculus I


(d) ( ) 3 7

5 2

29T x x xx

= + −

All of the terms in this function have roots in them. In order to use the power rule we need to first convert all the roots to fractional exponents. Again, remember that the Power Rule requires us to have a variable to a number and that it must be in the numerator of the term. Here is the function written in “proper” form.

( ) ( )( )

1 172 3

12 5

7132

25

7 213 52

29

29

9 2

T x x xx

x xx

x x x−

= + −

= + −

= + −

In the last two terms we combined the exponents. You should always do this with this kind of term. In a later section we will learn of a technique that would allow us to differentiate this term without combining exponents, however it will take significantly more work to do. Also don’t forget to move the term in the denominator of the third term up to the numerator. We can now differentiate the function.

( )

4 713 52

4 713 52

1 7 29 22 3 51 63 42 3 5

T x x x x

x x x

−−

−−

′ = + − −

= + +

Make sure that you can deal with fractional exponents. You will see a lot of them in this class.


(e) ( ) 2h x x xπ= −

In all of the previous examples the exponents have been nice integers or fractions. That is usually what we’ll see in this class. However, the exponent only needs to be a number so don’t get excited about problems like this one. They work exactly the same.

( ) 1 2 12h x x xππ − −′ = − The answer is a little messy and we won’t reduce the exponents down to decimals. However, this problem is not terribly difficult it just looks that way initially.

[Return to Problems] There is a general rule about derivatives in this class that you will need to get into the habit of using. When you see radicals you should always first convert the radical to a fractional exponent and then simplify exponents as much as possible. Following this rule will save you a lot of grief in the future.


Calculus I


Back when we first put down the properties we noted that we hadn’t included a property for products and quotients. That doesn’t mean that we can’t differentiate any product or quotient at this point. There are some that we can do. Example 2 Differentiate each of the following functions.

(a) ( )3 2 22y x x x= − [Solution]

(b) ( )5 2

2

2 5t th tt

+ −= [Solution]

Solution

(a) ( )3 2 22y x x x= −

In this function we can’t just differentiate the first term, differentiate the second term and then multiply the two back together. That just won’t work. We will discuss this in detail in the next section so if you’re not sure you believe that hold on for a bit and we’ll be looking at that soon as well as showing you an example of why it won’t work. It is still possible to do this derivative however. All that we need to do is convert the radical to fractional exponents (as we should anyway) and then multiply this through the parenthesis.

( )2 5 8

23 3 32 2y x x x x x= − = − Now we can differentiate the function.

2 53 310 8

3 3y x x′ = −


(b) ( )5 2

2

2 5t th tt

+ −=

As with the first part we can’t just differentiate the numerator and the denominator and the put it back together as a fraction. Again, if you’re not sure you believe this hold on until the next section and we’ll take a more detailed look at this. We can simplify this rational expression however as follows.

( )5 2

3 22 2 2

2 5 2 1 5t th t t tt t t

−= + − = + −

This is a function that we can differentiate. ( ) 2 36 10h t t t−′ = +



Calculus I


So, as we saw in this example there are a few products and quotients that we can differentiate. If we can first do some simplification the functions will sometimes simplify into a form that can be differentiated using the properties and formulas in this section. Before moving on to the next section let’s work a couple of examples to remind us once again of some of the interpretations of the derivative.

Example 3 Is ( ) 33

3002 4f x xx

= + + increasing, decreasing or not changing at 2x = − ?

Solution We know that the rate of change of a function is given by the functions derivative so all we need to do is it rewrite the function (to deal with the second term) and then take the derivative.

( ) ( )3 3 2 4 24

9002 300 4 6 900 6f x x x f x x x xx

− −′= + + ⇒ = − = −

Note that we rewrote the last term in the derivative back as a fraction. This is not something we’ve done to this point and is only being done here to help with the evaluation in the next step. It’s often easier to do the evaluation with positive exponents. So, upon evaluating the derivative we get

( ) ( ) 900 1292 6 4 32.2516 4

f ′ − = − = − = −

So, at 2x = − the derivative is negative and so the function is decreasing at 2x = − .

Example 4 Find the equation of the tangent line to ( ) 4 8f x x x= − at 16x = . Solution We know that the equation of a tangent line is given by, ( ) ( )( )y f a f a x a′= + − So, we will need the derivative of the function (don’t forget to get rid of the radical).

( ) ( )1 12 2

12

44 8 4 4 4f x x x f x xx

−′= − ⇒ = − = −

Again, notice that we eliminated the negative exponent in the derivative solely for the sake of the evaluation. All we need to do then is evaluate the function and the derivative at the point in question, 16x = .

( ) ( ) ( ) 416 64 8 4 32 4 34

f f x′= − = = − =

The tangent line is then, ( )32 3 16 3 16y x x= + − = −


Calculus I


Example 5 The position of an object at any time t (in hours) is given by, ( ) 3 22 21 60 10s t t t t= − + − Determine when the object is moving to the right and when the object is moving to the left. Solution The only way that we’ll know for sure which direction the object is moving is to have the velocity in hand. Recall that if the velocity is positive the object is moving off to the right and if the velocity is negative then the object is moving to the left. So, we need the derivative since the derivative is the velocity of the object. The derivative is,

( ) ( ) ( ) ( )2 26 42 60 6 7 10 6 2 5s t t t t t t t′ = − + = − + = − − The reason for factoring the derivative will be apparent shortly. Now, we need to determine where the derivative is positive and where the derivative is negative. There are several ways to do this. The method that I tend to prefer is the following. Since polynomials are continuous we know from the Intermediate Value Theorem that if the polynomial ever changes sign then it must have first gone through zero. So, if we knew where the derivative was zero we would know the only points where the derivative might change sign. We can see from the factored form of the derivative that the derivative will be zero at 2t = and

5t = . Let’s graph these points on a number line.

Now, we can see that these two points divide the number line into three distinct regions. In each of these regions we know that the derivative will be the same sign. Recall the derivative can only change sign at the two points that are used to divide the number line up into the regions. Therefore, all that we need to do is to check the derivative at a test point in each region and the derivative in that region will have the same sign as the test point. Here is the number line with the test points and results shown.


Calculus I


Here are the intervals in which the derivative is positive and negative.

positive : 2 & 5negative : 2 5

t tt

−∞ < < < < ∞< <

We included negative t’s here because we could even though they may not make much sense for this problem. Once we know this we also can answer the question. The object is moving to the right and left in the following intervals.

moving to the right : 2 & 5moving to the left : 2 5

t tt

−∞ < < < < ∞< <

Make sure that you can do the kind of work that we just did in this example. You will be asked numerous times over the course of the next two chapters to determine where functions are positive and/or negative. If you need some review or want to practice these kinds of problems you should check out the Solving Inequalities section of my Algebra/Trig Review.


Calculus I


Product and Quotient Rule In the previous section we noted that we had to be careful when differentiating products or quotients. It’s now time to look at products and quotients and see why. First let’s take a look at why we have to be careful with products and quotients. Suppose that we have the two functions ( ) 3f x x= and ( ) 6g x x= . Let’s start by computing the derivative of

the product of these two functions. This is easy enough to do directly.

( ) ( ) ( )3 6 9 89f g x x x x′ ′′ = = = Remember that on occasion we will drop the (x) part on the functions to simplify notation somewhat. We’ve done that in the work above. Now, let’s try the following.

( ) ( ) ( )( )2 5 73 6 18f x g x x x x′ ′ = =

So, we can very quickly see that.

( )f g f g′ ′ ′≠ In other words, the derivative of a product is not the product of the derivatives. Using the same functions we can do the same thing for quotients.

( )3

3 46 3 4

1 33f x x xg x x x

− −′′ ′ ′= = = = − = −

( )( )

2

5 3

3 16 2

f x xg x x x

′= =

′

So, again we can see that,

f fg g

′ ′ ≠ ′

To differentiate products and quotients we have the Product Rule and the Quotient Rule. Product Rule If the two functions f(x) and g(x) are differentiable (i.e. the derivative exist) then the product is differentiable and,

( )f g f g f g′ ′ ′= + The proof of the Product Rule is shown in the Proof of Various Derivative Formulas section of the Extras chapter.


Calculus I


Quotient Rule If the two functions f(x) and g(x) are differentiable (i.e. the derivative exist) then the quotient is differentiable and,

2

f f g f gg g

′ ′ ′ −=

Note that the numerator of the quotient rule is very similar to the product rule so be careful to not mix the two up! The proof of the Quotient Rule is shown in the Proof of Various Derivative Formulas section of the Extras chapter. Let’s do a couple of examples of the product rule. Example 1 Differentiate each of the following functions.

(a) ( )3 2 22y x x x= − [Solution]

(b) ( ) ( )( )36 10 20f x x x x= − − [Solution] Solution At this point there really aren’t a lot of reasons to use the product rule. As we noted in the previous section all we would need to do for either of these is to just multiply out the product and then differentiate. With that said we will use the product rule on these so we can see an example or two. As we add more functions to our repertoire and as the functions become more complicated the product rule will become more useful and in many cases required.

(a) ( )3 2 22y x x x= −

Note that we took the derivative of this function in the previous section and didn’t use the product rule at that point. We should however get the same result here as we did then. Now let’s do the problem here. There’s not really a lot to do here other than use the product rule. However, before doing that we should convert the radical to a fractional exponent as always.

( )2

23 2y x x x= − Now let’s take the derivative. So we take the derivative of the first function times the second then add on to that the first function times the derivative of the second function.

( ) ( )1 2

23 32 2 2 23

y x x x x x−

′ = − + −


Calculus I


This is NOT what we got in the previous section for this derivative. However, with some simplification we can arrive at the same answer.

2 5 2 5 2 53 3 3 3 3 34 2 10 82 2

3 3 3 3y x x x x x x′ = − + − = −

This is what we got for an answer in the previous section so that is a good check of the product rule.


(b) ( ) ( )( )36 10 20f x x x x= − −

This one is actually easier than the previous one. Let’s just run it through the product rule.

( ) ( )( ) ( )( )2 3

3 2

18 1 10 20 6 20

480 180 40 10

f x x x x x

x x x

′ = − − + − −

= − + + −

Since it was easy to do we went ahead and simplified the results a little.

[Return to Problems] Let’s now work an example or two with the quotient rule. In this case, unlike the product rule examples, a couple of these functions will require the quotient rule in order to get the derivative. The last two however, we can avoid the quotient rule if we’d like to as we’ll see. Example 2 Differentiate each of the following functions.

(a) ( ) 3 92zW z

z+

=−

[Solution]

(b) ( ) 2

42

xh xx

=−

[Solution]

(c) ( ) 6

4f xx

= [Solution]

(d) 6

5wy = [Solution]

Solution

(a) ( ) 3 92zW z

z+

=−

There isn’t a lot to do here other than to use the quotient rule. Here is the work for this function.

( ) ( ) ( ) ( )

( )

( )

2

2

3 2 3 9 12

152

z zW z

z

z

− − + −′ =

−

=−



Calculus I


(b) ( ) 2

42

xh xx

=−

Again, not much to do here other than use the quotient rule. Don’t forget to convert the square root into a fractional exponent.

( )( ) ( ) ( )

( )

( )

( )

1 121 2 2

222

3 1 32 2 2

22

3 12 2

22

4 2 4 2

2

2 4 8

2

6 4

2

x x x xh x

x

x x x

x

x x

x

−

−

−

− −′ =

−

− −=

−

− −=

−


(c) ( ) 6

4f xx

=

It seems strange to have this one here rather than being the first part of this example given that it definitely appears to be easier than any of the previous two. In fact, it is easier. There is a point to doing it here rather than first. In this case there are two ways to do compute this derivative. There is an easy way and a hard way and in this case the hard way is the quotient rule. That’s the point of this example. Let’s do the quotient rule and see what we get.

( )( )( ) ( )

( )

6 5 5

2 12 76

0 4 6 24 24x x xf xx xx

− −′ = = = −

Now, that was the “hard” way. So, what was so hard about it? Well actually it wasn’t that hard, there is just an easier way to do it that’s all. However, having said that, a common mistake here is to do the derivative of the numerator (a constant) incorrectly. For some reason many people will give the derivative of the numerator in these kinds of problems as a 1 instead of 0! Also, there is some simplification that needs to be done in these kinds of problems if you do the quotient rule. The easy way is to do what we did in the previous section.

( ) 6 77

244 24f x x xx

− −′ = = − = −

Either way will work, but I’d rather take the easier route if I had the choice.



Calculus I


(d) 6

5wy =

This problem also seems a little out of place. However, it is here again to make a point. Do not confuse this with a quotient rule problem. While you can do the quotient rule on this function there is no reason to use the quotient rule on this. Simply rewrite the function as

615

y w=

and differentiate as always. 56

5y w′ =

[Return to Problems] Finally, let’s not forget about our applications of derivatives. Example 3 Suppose that the amount of air in a balloon at any time t is given by

( )36

4 1tV t

t=

+

Determine if the balloon is being filled with air or being drained of air at 8t = . Solution If the balloon is being filled with air then the volume is increasing and if it’s being drained of air then the volume will be decreasing. In other words, we need to get the derivative so that we can determine the rate of change of the volume at 8t = . This will require the quotient rule.

( ) ( ) ( )( )

( )

( )

2 13 3

2

1 23 3

2

13

23

2

2 4 1 6 44 1

16 24 1

216

4 1

t t tV t

t

t tt

tt

t

−

−

+ −′ =

+

− +=

+

− +

=+

Note that we simplified the numerator more than usual here. This was only done to make the derivative easier to evaluate. The rate of change of the volume at 8t = is then,


Calculus I


( )

( )

( )( ) ( ) ( ) ( )

21 2 1 23 3 3

2

216 248 8 2 8 8 2 4

3363 7

2178 242

V− + ′ = = = = =

= − = −

So, the rate of change of the volume at 8t = is negative and so the volume must be decreasing. Therefore air is being drained out of the balloon at 8t = . As a final topic let’s note that the product rule can be extended to more than two functions, for instance.

( )

( )

f g h f g h f g h f g h

f g h w f g h w f g h w f g h w f g h w

′ ′ ′ ′= + +

′ ′ ′ ′ ′= + + +

With this section and the previous section we are now able to differentiate powers of x as well as sums, differences, products and quotients of these kinds of functions. However, there are many more functions out there in the world that are not in this form. The next few sections give many of these functions as well as give their derivatives.


Calculus I


Derivatives of Trig Functions With this section we’re going to start looking at the derivatives of functions other than polynomials or roots of polynomials. We’ll start this process off by taking a look at the derivatives of the six trig functions. Two of the derivatives will be derived. The remaining four are left to the reader and will follow similar proofs for the two given here. Before we actually get into the derivatives of the trig functions we need to give a couple of limits that will show up in the derivation of two of the derivatives. Fact

0 0

sin cos 1lim 1 lim 0θ θ

θ θθ θ→ →

−= =

See the Proof of Trig Limits section of the Extras chapter to see the proof of these two limits. Before we start differentiating trig functions let’s work a quick set of limit problems that this fact now allows us to do. Example 1 Evaluate each of the following limits.

(a) 0

sinlim6θ

θθ→

[Solution]

(b) ( )0

sin 6limx

xx→

[Solution]

(c) ( )0

limsin 7x

xx→

[Solution]

(d) ( )( )0

sin 3lim

sin 8t

tt→

[Solution]

(e) ( )4

sin 4lim

4x

xx→

−−

[Solution]

(f) ( )0

cos 2 1limz

zz→

− [Solution]

Solution

(a) 0

sinlim6θ

θθ→

There really isn’t a whole lot to this limit. In fact, it’s only here to contrast with the next example so you can see the difference in how these work. In this case since there is only a 6 in the denominator we’ll just factor this out and then use the fact.

( )0 0

sin 1 sin 1 1lim lim 16 6 6 6θ θ

θ θθ θ→ →

= = =



Calculus I


(b) ( )

0

sin 6limx

xx→

Now, in this case we can’t factor the 6 out of the sine so we’re stuck with it there and we’ll need to figure out a way to deal with it. To do this problem we need to notice that in the fact the argument of the sine is the same as the denominator (i.e. both θ ’s). So we need to get both of the argument of the sine and the denominator to be the same. We can do this by multiplying the numerator and the denominator by 6 as follows.

( ) ( ) ( )

0 0 0

sin 6 6sin 6 sin 6lim lim 6lim

6 6x x x

x x xx x x→ → →

= =

Note that we factored the 6 in the numerator out of the limit. At this point, while it may not look like it, we can use the fact above to finish the limit. To see that we can use the fact on this limit let’s do a change of variables. A change of variables is really just a renaming of portions of the problem to make something look more like something we know how to deal with. They can’t always be done, but sometimes, such as this case, they can simplify the problem. The change of variables here is to let 6xθ = and then notice that as

0x → we also have ( )6 0 0θ → = . When doing a change of variables in a limit we need to

change all the x’s into θ ’s and that includes the one in the limit. Doing the change of variables on this limit gives,

( ) ( )

( )

( )

0 0

0

sin 6 sin 6lim 6lim let 6

6sin

6lim

6 16

x x

x xx

x x

θ

θ

θθ

→ →

→

= =

=

=

=

And there we are. Note that we didn’t really need to do a change of variables here. All we really need to notice is that the argument of the sine is the same as the denominator and then we can use the fact. A change of variables, in this case, is really only needed to make it clear that the fact does work.


(c) ( )0

limsin 7x

xx→

In this case we appear to have a small problem in that the function we’re taking the limit of here is upside down compared to that in the fact. This is not the problem it appears to be once we notice that,


Calculus I


( ) ( )

1sin 7sin 7

xxx

x

=

and then all we need to do is recall a nice property of limits that allows us to do ,

( ) ( )

( )

( )

0 0

0

0

0

1lim limsin 7sin 7

lim1

sin 7lim

1sin 7

lim

x x

x

x

x

xxx

x

xx

xx

→ →

→

→

→

=

=

=

With a little rewriting we can see that we do in fact end up needing to do a limit like the one we did in the previous part. So, let’s do the limit here and this time we won’t bother with a change of variable to help us out. All we need to do is multiply the numerator and denominator of the fraction in the denominator by 7 to get things set up to use the fact. Here is the work for this limit.

( ) ( )

( )

( ) ( )

0

0

0

1lim7sin 7sin 7 lim

71

sin 77 lim

71

7 117

x

x

x

xxx

x

xx

→

→

→

=

=

=

=


(d) ( )( )0

sin 3lim

sin 8t

tt→

This limit looks nothing like the limit in the fact, however it can be thought of as a combination of the previous two parts by doing a little rewriting. First, we’ll split the fraction up as follows,

( )( )

( )( )0 0

sin 3 sin 3 1lim limsin 8 1 sin 8t t

t tt t→ →

=

Now, the fact wants a t in the denominator of the first and in the numerator of the second. This is


Calculus I


easy enough to do if we multiply the whole thing by tt (which is just one after all and so won’t

change the problem) and then do a little rearranging as follows,

( )( )

( )( )

( )( )

( )( )

0 0

0

0 0

sin 3 sin 3 1lim limsin 8 1 sin 8

sin 3lim

sin 8

sin 3lim lim

sin 8

t t

t

t t

t t tt t t

t tt t

t tt t

→ →

→

→ →

=

=

=

At this point we can see that this really is two limits that we’ve seen before. Here is the work for each of these and notice on the second limit that we’re going to work it a little differently than we did in the previous part. This time we’re going to notice that it doesn’t really matter whether the sine is in the numerator or the denominator as long as the argument of the sine is the same as what’s in the numerator the limit is still one. Here is the work for this limit.

( )( )

( )( )

( )( )

( )

0 0 0

0 0

sin 3 3sin 3 8lim lim limsin 8 3 8sin 8

sin 3 1 83lim lim3 8 sin 8

138

38

t t t

t t

t t tt t t

t tt t

→ → →

→ →

=

=

=

=


(e) ( )

4

sin 4lim

4x

xx→

−−

This limit almost looks the same as that in the fact in the sense that the argument of the sine is the same as what is in the denominator. However, notice that, in the limit, x is going to 4 and not 0 as the fact requires. However, with a change of variables we can see that this limit is in fact set to use the fact above regardless. So, let xθ = − 4 and then notice that as 4x → we have 0θ → . Therefore, after doing the change of variable the limit becomes,

( )

4 0

sin 4 sinlim lim 14x

xx θ

θθ→ →

−= =

−



Calculus I


(f) ( )

0

cos 2 1limz

zz→

−

The previous parts of this example all used the sine portion of the fact. However, we could just have easily used the cosine portion so here is a quick example using the cosine portion to illustrate this. We’ll not put in much explanation here as this really does work in the same manner as the sine portion.

( ) ( )( )

( )

( )

0 0

0

2 cos 2 1cos 2 1lim lim

2cos 2 1

2lim2

2 00

z z

z

zzz z

zz

→ →

→

−−=

−=

=

All that is required to use the fact is that the argument of the cosine is the same as the denominator.

[Return to Problems] Okay, now that we’ve gotten this set of limit examples out of the way let’s get back to the main point of this section, differentiating trig functions. We’ll start with finding the derivative of the sine function. To do this we will need to use the definition of the derivative. It’s been a while since we’ve had to use this, but sometimes there just isn’t anything we can do about it. Here is the definition of the derivative for the sine function.

( )( ) ( ) ( )0

sin sinsin lim

h

x h xd xdx h→

+ −=

Since we can’t just plug in 0h = to evaluate the limit we will need to use the following trig formula on the first sine in the numerator. ( ) ( ) ( ) ( ) ( )sin sin cos cos sinx h x h x h+ = + Doing this gives us,

( )( ) ( ) ( ) ( ) ( ) ( )

( ) ( )( ) ( ) ( )

( ) ( ) ( ) ( )

0

0

0 0

sin cos cos sin sinsin lim

sin cos 1 cos sinlim

cos 1 sinlimsin limcos

h

h

h h

x h x h xd xdx h

x h x hh

h hx x

h h

→

→

→ →

+ −=

− +=

−= +


Calculus I


As you can see upon using the trig formula we can combine the first and third term and then factor a sine out of that. We can then break up the fraction into two pieces, both of which can be dealt with separately. Now, both of the limits here are limits as h approaches zero. In the first limit we have a sin(x) and in the second limit we have a cos(x). Both of these are only functions of x only and as h moves in towards zero this has no affect on the value of x. Therefore, as far as the limits are concerned, these two functions are constants and can be factored out of their respective limits. Doing this gives,

( )( ) ( ) ( ) ( ) ( )0 0

cos 1 sinsin sin lim cos lim

h h

h hd x x xdx h h→ →

−= +

At this point all we need to do is use the limits in the fact above to finish out this problem.

( )( ) ( )( ) ( ) ( ) ( )sin sin 0 cos 1 cosd x x x xdx

= + =

Differentiating cosine is done in a similar fashion. It will require a different trig formula, but other than that is an almost identical proof. The details will be left to you. When done with the proof you should get,

( )( ) ( )cos sind x xdx

= −

With these two out of the way the remaining four are fairly simple to get. All the remaining four trig functions can be defined in terms of sine and cosine and these definitions, along with appropriate derivative rules, can be used to get their derivatives. Let’s take a look at tangent. Tangent is defined as,

( ) ( )( )

sintan

cosx

xx

=

Now that we have the derivatives of sine and cosine all that we need to do is use the quotient rule on this. Let’s do that.

( )( ) ( )( )

( ) ( ) ( ) ( )( )( )( )

( ) ( )( )

2

2 2

2

sintan

cos

cos cos sin sin

cos

cos sincos

xd dxdx dx x

x x x x

x

x xx

=

− −

=

+=


Calculus I


Now, recall that ( ) ( )2 2cos sin 1x x+ = and if we also recall the definition of secant in terms of

cosine we arrive at,

( )( ) ( ) ( )( )

( )( )

2 2

2

2

2

cos sintan

cos1

cos

sec

x xd xdx x

x

x

+=

=

=

The remaining three trig functions are also quotients involving sine and/or cosine and so can be differentiated in a similar manner. We’ll leave the details to you. Here are the derivatives of all six of the trig functions. Derivatives of the six trig functions

( )( ) ( ) ( )( ) ( )

( )( ) ( ) ( )( ) ( )

( )( ) ( ) ( ) ( )( ) ( ) ( )

2 2

sin cos cos sin

tan sec cot csc

sec sec tan csc csc cot

d dx x x xdx dxd dx x x xdx dxd dx x x x x xdx dx

= = −

= = −

= = −

At this point we should work some examples. Example 2 Differentiate each of the following functions.

(a) ( ) ( ) ( )3sec 10cotg x x x= − [Solution]

(b) ( ) ( )4 23 tanh w w w w−= − [Solution]

(c) ( ) ( ) ( )5sin cos 4cscy x x x= + [Solution]

(d) ( ) ( )( )

sin3 2cos

tP t

t=

− [Solution]

Solution (a) ( ) ( ) ( )3sec 10cotg x x x= −

There really isn’t a whole lot to this problem. We’ll just differentiate each term using the formulas from above.

( ) ( ) ( ) ( )( )

( ) ( ) ( )

2

2

3sec tan 10 csc

3sec tan 10csc

g x x x x

x x x

′ = − −

= +


(b) ( ) ( )4 23 tanh w w w w−= −

In this part we will need to use the product rule on the second term and note that we really will need the product rule here. There is no other way to do this derivative unlike what we saw when


Calculus I


we first looked at the product rule. When we first looked at the product rule the only functions we knew how to differentiate were polynomials and in those cases all we really needed to do was multiply them out and we could take the derivative without the product rule. We are now getting into the point where we will be forced to do the product rule at times regardless of whether or not we want to. We will also need to be careful with the minus sign in front of the second term and make sure that it gets dealt with properly. There are two ways to deal with this. One way it to make sure that you use a set of parenthesis as follows,

( ) ( ) ( )( )

( ) ( )

5 2 2

5 2 2

12 2 tan sec

12 2 tan sec

h w w w w w w

w w w w w

−

−

′ = − − +

= − − −

Because the second term is being subtracted off of the first term then the whole derivative of the second term must also be subtracted off of the derivative of the first term. The parenthesis make this idea clear. A potentially easier way to do this is to think of the minus sign as part of the first function in the product. Or, in other words the two functions in the product, using this idea, are 2w− and

( )tan w . Doing this gives,

( ) ( ) ( )5 2 212 2 tan sech w w w w w w−′ = − − − So, regardless of how you approach this problem you will get the same derivative.

[Return to Problems] (c) ( ) ( ) ( )5sin cos 4cscy x x x= +

As with the previous part we’ll need to use the product rule on the first term. We will also think of the 5 as part of the first function in the product to make sure we deal with it correctly. Alternatively, you could make use of a set of parenthesis to make sure the 5 gets dealt with properly. Either way will work, but we’ll stick with thinking of the 5 as part of the first term in the product. Here’s the derivative of this function.

( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( )2 2

5cos cos 5sin sin 4csc cot

5cos 5sin 4csc cot

y x x x x x x

x x x x

′ = + − −

= − −


(d) ( ) ( )( )

sin3 2cos

tP t

t=

−

In this part we’ll need to use the quotient rule to take the derivative.


Calculus I


( ) ( ) ( )( ) ( ) ( )( )( )( )

( ) ( ) ( )( )( )

2

2 2

2

cos 3 2cos sin 2sin

3 2cos

3cos 2cos 2sin

3 2cos

t t t tP t

t

t t t

t

− −′ =

−

− −=

−

Be careful with the signs when differentiating the denominator. The negative sign we get from differentiating the cosine will cancel against the negative sign that is already there. This appears to be done, but there is actually a fair amount of simplification that can yet be done. To do this we need to factor out a “-2” from the last two terms in the numerator and the make use of the fact that ( ) ( )2 2cos sin 1θ θ+ = .

( )( ) ( ) ( )( )

( )( )( )

( )( )

2 2

2

2

3cos 2 cos sin

3 2cos

3cos 2

3 2cos

t t tP t

t

t

t

− +′ =

−

−=

−

[Return to Problems] As a final problem here let’s not forget that we still have our standard interpretations to derivatives. Example 3 Suppose that the amount of money in a bank account is given by ( ) ( ) ( )500 100cos 150sinP t t t= + − where t is in years. During the first 10 years in which the account is open when is the amount of money in the account increasing? Solution To determine when the amount of money is increasing we need to determine when the rate of change is positive. Since we know that the rate of change is given by the derivative that is the first thing that we need to find. ( ) ( ) ( )100sin 150cosP t t t′ = − − Now, we need to determine where in the first 10 years this will be positive. This is equivalent to asking where in the interval [0, 10] is the derivative positive. Recall that both sine and cosine are continuous functions and so the derivative is also a continuous function. The Intermediate Value Theorem then tells us that the derivative can only change sign if it first goes through zero. So, we need to solve the following equation.


Calculus I


( ) ( )( ) ( )( )( )( )

100sin 150cos 0

100sin 150cos

sin1.5

cos

tan 1.5

t t

t t

tt

t

− − =

= −

= −

= −

The solution to this equation is,

2.1588 2 , 0, 1, 2,5.3004 2 , 0, 1, 2,

t n nt n n

ππ

= + = ± ±= + = ± ±

……

If you don’t recall how to solve trig equations go back and take a look at the sections on solving trig equations in the Review chapter. We are only interested in those solutions that fall in the range [0, 10]. Plugging in values of n into the solutions above we see that the values we need are,

2.1588 2.1588 2 8.44205.3004

t tt

π= = + ==

So, much like solving polynomial inequalities all that we need to do is sketch in a number line and add in these points. These points will divide the number line into regions in which the derivative must always be the same sign. All that we need to do then is choose a test point from each region to determine the sign of the derivative in that region. Here is the number line with all the information on it.

So, it looks like the amount of money in the bank account will be increasing during the following intervals. 2.1588 5.3004 8.4420 10t t< < < < Note that we can’t say anything about what is happening after 10t = since we haven’t done any work for t’s after that point. In this section we saw how to differentiate trig functions. We also saw in the last example that our interpretations of the derivative are still valid so we can’t forget those.


Calculus I


Also, it is important that we be able to solve trig equations as this is something that will arise off and on in this course. It is also important that we can do the kinds of number lines that we used in the last example to determine where a function is positive and where a function is negative. This is something that we will be doing on occasion in both this chapter and the next.


Calculus I


Derivatives of Exponential and Logarithm Functions The next set of functions that we want to take a look at are exponential and logarithm functions. The most common exponential and logarithm functions in a calculus course are the natural exponential function, xe , and the natural logarithm function, ( )ln x . We will take a more

general approach however and look at the general exponential and logarithm function. Exponential Functions We’ll start off by looking at the exponential function,

( ) xf x a=

We want to differentiate this. The power rule that we looked at a couple of sections ago won’t work as that required the exponent to be a fixed number and the base to be a variable. That is exactly the opposite from what we’ve got with this function. So, we’re going to have to start with the definition of the derivative.

( ) ( ) ( )

( )

0

0

0

0

lim

lim

lim

1lim

h

x h x

h

x h x

h

x h

h

f x h f xf x

ha a

ha a a

ha a

h

→

+

→

→

→

+ −′ =

−=

−=

−=

Now, the xa is not affected by the limit since it doesn’t have any h’s in it and so is a constant as far as the limit is concerned. We can therefore factor this out of the limit. This gives,

( )0

1limh

x

h

af x ah→

−′ =

Now let’s notice that the limit we’ve got above is exactly the definition of the derivative of

( ) xf x a= at 0x = , i.e. ( )0f ′ . Therefore, the derivative becomes,

( ) ( )0 xf x f a′ ′= So, we are kind of stuck. We need to know the derivative in order to get the derivative! There is one value of a that we can deal with at this point. Back in the Exponential Functions section of the Review chapter we stated that 2.71828182845905=e … What we didn’t do however is actually define where e comes from. There are in fact a variety of ways to define e. Here are a three of them.


Calculus I


Some Definitions of e.

1. 1lim 1

n

n n→∞

= +

e

2. e is the unique positive number for which 0

1lim 1h

h

h→

−=

e

3. 0

1!n n

∞

=

= ∑e

The second one is the important one for us because that limit is exactly the limit that we’re working with above. So, this definition leads to the following fact, Fact 1

For the natural exponential function, ( ) xf x = e we have ( )0

10 lim 1h

hf

h→

−′ = =e

.

So, provided we are using the natural exponential function we get the following. ( ) ( )x xf x f x′= ⇒ =e e At this point we’re missing some knowledge that will allow us to easily get the derivative for a general function. Eventually we will be able to show that for a general exponential function we have, ( ) ( ) ( )lnx xf x a f x a a′= ⇒ = Logarithm Functions Let’s now briefly get the derivatives for logarithms. In this case we will need to start with the following fact about functions that are inverses of each other. Fact 2 If f(x) and g(x) are inverses of each other then,

( ) ( )( )1g x

f g x′ =

′

So, how is this fact useful to us? Well recall that the natural exponential function and the natural logarithm function are inverses of each other and we know what the derivative of the natural exponential function is! So, if we have ( ) xf x = e and ( ) lng x x= then,


Calculus I


( ) ( )( ) ( ) ln

1 1 1 1xg xg x

xf g x′ = = = =

′ ee

The last step just uses the fact that the two functions are inverses of each other. Putting this all together gives,

( ) 1ln 0d x xdx x

= >

Note that we need to require that 0x > since this is required for the logarithm and so must also be required for its derivative. It can also be shown that,

( ) 1ln 0d x xdx x

= ≠

Using this all we need to avoid is 0x = . In this case, unlike the exponential function case, we can actually find the derivative of the general logarithm function. All that we need is the derivative of the natural logarithm, which we just found, and the change of base formula. Using the change of base formula we can write a general logarithm as,

lnloglna

xxa

=

Differentiation is then fairly simple.

( )

( )

lnlogln

1 lnln

1ln

ad d xxdx dx a

d xa dx

x a

=

=

=

We took advantage of the fact that a was a constant and so ln a is also a constant and can be factored out of the derivative. Putting all this together gives,

( ) 1loglna

d xdx x a

=

Here is a summary of the derivatives in this section.

( ) ( )

( ) ( )

ln

1 1ln logln

x x x x

a

d d a a adx dxd dx xdx x dx x a

= =

= =

e e


Calculus I


Okay, now that we have the derivations of the formulas out of the way let’s compute a couple of derivatives. Example 1 Differentiate each of the following functions.

(a) ( ) 94 5logwR w w= −

(b) ( ) 33 10 lnxf x x x= +e

(c) 53 1

x

xy =+

ee

Solution (a) This will be the only example that doesn’t involve the natural exponential and natural logarithm functions.

( ) 54 ln 4ln 9

wR ww

′ = −

(b) Not much to this one. Just remember to use the product rule on the second term.

( ) 2 3

2 2

13 30 ln 10

3 30 ln 10

x

x

f x x x xx

x x x

′ = + +

= + +

e

e

(c) We’ll need to use the quotient rule on this one.

( ) ( )( )( )

( )

( )

2

2 2

2

2

5 3 1 5 3

3 1

15 5 15

3 1

5

3 1

x x x x

x

x x x

x

x

x

y+ −

=+

+ −=

+

=+

e e e e

e

e e e

e

e

e

There’s really not a lot to differentiating natural logarithms and natural exponential functions at this point as long as you remember the formulas. In later sections as we get more formulas under our belt they will become more complicated. Next, we need to do our obligatory application/interpretation problem so we don’t forget about them. Example 2 Suppose that the position of an object is given by ( ) ts t t= e Does the object ever stop moving? Solution First we will need the derivative. We need this to determine if the object ever stops moving since


Calculus I


at that point (provided there is one) the velocity will be zero and recall that the derivative of the position function is the velocity of the object. The derivative is, ( ) ( )1t t ts t t t′ = + = +e e e So, we need to determine if the derivative is ever zero. To do this we will need to solve, ( )1 0tt+ =e Now, we know that exponential functions are never zero and so this will only be zero at 1t = − . So, if we are going to allow negative values of t then the object will stop moving once at 1t = − . If we aren’t going to allow negative values of t then the object will never stop moving. Before moving on to the next section we need to go back over a couple of derivatives to make sure that we don’t confuse the two. The two derivatives are,

( )

( )

1 Power Rule

ln Derivative of an exponential function

n n

x x

d x nxdxd a a adx

−=

=

It is important to note that with the Power rule the exponent MUST be a constant and the base MUST be a variable while we need exactly the opposite for the derivative of an exponential function. For an exponential function the exponent MUST be a variable and the base MUST be a constant. It is easy to get locked into one of these formulas and just use it for both of these. We also haven’t even talked about what to do if both the exponent and the base involve variables. We’ll see this situation in a later section.


Calculus I


Derivatives of Inverse Trig Functions In this section we are going to look at the derivatives of the inverse trig functions. In order to derive the derivatives of inverse trig functions we’ll need the formula from the last section relating the derivatives of inverse functions. If f(x) and g(x) are inverse functions then,

( ) ( )( )1g x

f g x′ =

′

Recall as well that two functions are inverses if ( )( )f g x x= and ( )( )g f x x= .

We’ll go through inverse sine, inverse cosine and inverse tangent in detail here and leave the other three to you to derive if you’d like to. Inverse Sine Let’s start with inverse sine. Here is the definition of the inverse sine.

1sin sin for2 2

y x y x yπ π−= ⇔ = − ≤ ≤

So, evaluating an inverse trig function is the same as asking what angle (i.e. y) did we plug into the sine function to get x. The restrictions on y given above are there to make sure that we get a consistent answer out of the inverse sine. We know that there are in fact an infinite number of angles that will work and we want a consistent value when we work with inverse sine. Using the range of angles above gives all possible values of the sine function exactly once. If you’re not sure of that sketch out a unit circle and you’ll see that that range of angles (the y’s) will cover all possible values of sine. Note as well that since ( )1 sin 1y− ≤ ≤ we also have 1 1x− ≤ ≤ .

Let’s work a quick example.

Example 1 Evaluate 1 1sin2

−

Solution So we are really asking what angle y solves the following equation.

( ) 1sin2

y =

and we are restricted to the values of y above.

From a unit circle we can quickly see that 6

y π= .

We have the following relationship between the inverse sine function and the sine function.


Calculus I


( ) ( )1 1sin sin sin sinx x x x− −= = In other words they are inverses of each other. This means that we can use the fact above to find the derivative of inverse sine. Let’s start with, ( ) ( ) 1sin sinf x x g x x−= = Then,

( ) ( )( ) ( )1

1 1cos sin

g xf g x x−

′ = =′

This is not a very useful formula. Let’s see if we can get a better formula. Let’s start by recalling the definition of the inverse sine function. ( ) ( )1sin siny x x y−= ⇒ = Using the first part of this definition the denominator in the derivative becomes,

( ) ( )1cos sin cosx y− = Now, recall that

2 2 2cos sin 1 cos 1 siny y y y+ = ⇒ = − Using this, the denominator is now,

( ) ( )1 2cos sin cos 1 sinx y y− = = − Now, use the second part of the definition of the inverse sine function. The denominator is then,

( )1 2 2cos sin 1 sin 1x y x− = − = − Putting all of this together gives the following derivative.

( )1

2

1sin1

d xdx x

− =−

Inverse Cosine Now let’s take a look at the inverse cosine. Here is the definition for the inverse cosine. 1cos cos for 0y x y x y π−= ⇔ = ≤ ≤ As with the inverse since we’ve got a restriction on the angles, y, that we get out of the inverse cosine function. Again, if you’d like to verify this a quick sketch of a unit circle should convince you that this range will cover all possible values of cosine exactly once. Also, we also have

1 1x− ≤ ≤ because ( )1 cos 1y− ≤ ≤ .


Calculus I


Example 2 Evaluate 1 2cos2

− −

.

Solution As with the inverse sine we are really just asking the following.

2cos2

y = −

where y must meet the requirements given above. From a unit circle we can see that we must

have 34

y π= .

The inverse cosine and cosine functions are also inverses of each other and so we have, ( ) ( )1 1cos cos cos cosx x x x− −= = To find the derivative we’ll do the same kind of work that we did with the inverse sine above. If we start with ( ) ( ) 1cos cosf x x g x x−= = then,

( ) ( )( ) ( )1

1 1sin cos

g xf g x x−

′ = =′ −

Simplifying the denominator here is almost identical to the work we did for the inverse sine and so isn’t shown here. Upon simplifying we get the following derivative.

( )1

2

1cos1

d xdx x

− = −−

So, the derivative of the inverse cosine is nearly identical to the derivative of the inverse sine. The only difference is the negative sign. Inverse Tangent Here is the definition of the inverse tangent.

1tan tan for2 2

y x y x yπ π−= ⇔ = − < <

Again, we have a restriction on y, but notice that we can’t let y be either of the two endpoints in the restriction above since tangent isn’t even defined at those two points. To convince yourself that this range will cover all possible values of tangent do a quick sketch of the tangent function and we can see that in this range we do indeed cover all possible values of tangent. Also, in this case there are no restrictions on x because tangent can take on all possible values.


Calculus I


Example 3 Evaluate 1tan 1− Solution Here we are asking, tan 1y =

where y satisfies the restrictions given above. From a unit circle we can see that 4

y π= .

Because there is no restriction on x we can ask for the limits of the inverse tangent function as x goes to plus or minus infinity. To do this we’ll need the graph of the inverse tangent function. This is shown below.

From this graph we can see that

1 1lim tan lim tan2 2x x

x xπ π− −

→∞ →−∞= = −

The tangent and inverse tangent functions are inverse functions so,

( ) ( )1 1tan tan tan tanx x x x− −= = Therefore to find the derivative of the inverse tangent function we can start with ( ) ( ) 1tan tanf x x g x x−= = We then have,

( ) ( )( ) ( )2 1

1 1sec tan

g xf g x x−

′ = =′

Simplifying the denominator is similar to the inverse sine, but different enough to warrant showing the details. We’ll start with the definition of the inverse tangent. 1tan tany x y x−= ⇒ = The denominator is then,

( )2 1 2sec tan secx y− =


Calculus I


Now, if we start with the fact that 2 2cos sin 1y y+ = and divide every term by cos2 y we will get, 2 21 tan secy y+ = The denominator is then,

( )2 1 2 2sec tan sec 1 tanx y y− = = + Finally using the second portion of the definition of the inverse tangent function gives us,

( )2 1 2 2sec tan 1 tan 1x y x− = + = + The derivative of the inverse tangent is then,

( )12

1tan1

d xdx x

− =+

There are three more inverse trig functions but the three shown here the most common ones. Formulas for the remaining three could be derived by a similar process as we did those above. Here are the derivatives of all six inverse trig functions.

( ) ( )

( ) ( )

( ) ( )

1 1

2 2

1 12 2

1 1

2 2

1 1sin cos1 11 1tan cot

1 11 1sec csc

1 1

d dx xdx dxx xd dx xdx x dx xd dx xdx dxx x x x

− −

− −

− −

= = −− −

= = −+ +

= = −− −

We should probably now do a couple of quick derivatives here before moving on to the next section. Example 4 Differentiate the following functions.

(a) ( ) ( ) ( )1 14cos 10 tanf t t t− −= −

(b) ( )1siny z z−= Solution (a) Not much to do with this one other than differentiate each term.

( ) 22

4 1011

f ttt

′ = − −+−


Calculus I


(b) Don’t forget to convert the radical to fractional exponents before using the product rule.

( )1

122

1 sin2 1

zy z zz

− −′ = +−

Alternate Notation There is some alternate notation that is used on occasion to denote the inverse trig functions. This notation is,

1 1

1 1

1 1

sin arcsin cos arccostan arctan cot arccot sec arcsec csc arccsc

x x x xx x x xx x x x

− −

− −

− −

= =

= =

= =


Calculus I


Derivatives of Hyperbolic Functions The last set of functions that we’re going to be looking in this chapter at are the hyperbolic functions. In many physical situations combinations of xe and x−e arise fairly often. Because of this these combinations are given names. There are six hyperbolic functions and they are defined as follows.

sinh cosh2 2

sinh cosh 1tanh cothcosh sinh tanh

1 1sech cschcosh sinh

x x x x

x x

x xx xx x x

x xx x

− −− += =

= = =

= =

e e e e

Here are the graphs of the three main hyperbolic functions.

We also have the following facts about the hyperbolic functions.

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )2 2 2 2

sinh sinh cosh cosh

cosh sinh 1 1 tanh sech

x x x x

x x x x

− = − − =

− = − =

You’ll note that these are similar, but not quite the same, to some of the more common trig identities so be careful to not confuse the identities here with those of the standard trig functions.


Calculus I


Because the hyperbolic functions are defined in terms of exponential functions finding their derivatives is fairly simple provided you’ve already read through the next section. We haven’t however so we’ll need the following formula that can be easily proved after we’ve covered the next section.

( )x xddx

− −= −e e

With this formula we’ll do the derivative for hyperbolic sine and leave the rest to you as an exercise.

( ) ( )sinh cosh

2 2 2

x xx x x xd dx xdx dx

−− −− − − += = = =

e ee e e e

For the rest we can either use the definition of the hyperbolic function and/or the quotient rule. Here are all six derivatives.

( ) ( )

( ) ( )

( ) ( )

2 2

sinh cosh cosh sinh

tanh sech coth csch

sech sech tanh csch csch coth

d dx x x xdx dxd dx x x xdx dxd dx x x x x xdx dx

= =

= = −

= − = −

Here are a couple of quick derivatives using hyperbolic functions. Example 1 Differentiate each of the following functions.

(a) ( ) 52 coshf x x x=

(b) ( ) sinh1th t

t=

+

Solution (a) ( ) 4 510 cosh 2 sinhf x x x x x′ = + (b)

( ) ( )( )2

1 cosh sinh1

t t th t

t+ −

′ =+


Calculus I


Chain Rule We’ve taken a lot of derivatives over the course of the last few sections. However, if you look back they have all been functions similar to the following kinds of functions.

( ) ( ) ( ) ( ) ( )50 tan lnwR z z f t t y x h w g x x= = = = =e These are all fairly simple functions in that wherever the variable appears it is by itself. What about functions like the following,

( ) ( ) ( )( ) ( )( )

( ) ( ) ( )4 2

50 33 2

3 9 4 4

5 8 2 cos tan 3 tan 5

lnw w

R z z f t t t y x x

h w g x x x− + −

= − = + = +

= = +e

None of our rules will work on these functions and yet some of these functions are closer to the derivatives that we’re liable to run into than the functions in the first set. Let’s take the first one for example. Back in the section on the definition of the derivative we actually used the definition to compute this derivative. In that section we found that,

( ) 52 5 8

R zz

′ =−

If we were to just use the power rule on this we would get,

( )12

1 15 82 2 5 8

zz

−− =−

which is not the derivative that we computed using the definition. It is close, but it’s not the same. So, the power rule alone simply won’t work to get the derivative here. Let’s keep looking at this function and note that if we define,

( ) ( ) 5 8f z z g z z= = − then we can write the function as a composition.

( ) ( ) ( ) ( )( ) 5 8R z f g z f g z z= = = −o and it turns out that it’s actually fairly simple to differentiate a function composition using the Chain Rule. There are two forms of the chain rule. Here they are. Chain Rule Suppose that we have two functions f(x) and g(x) and they are both differentiable.

1. If we define ( ) ( ) ( )F x f g x= o then the derivative of F(x) is,

( ) ( )( ) ( )F x f g x g x′ ′ ′=

2. If we have ( )y f u= and ( )u g x= then the derivative of y is,

dy dy dudx du dx

=


Calculus I


Each of these forms have their uses, however we will work mostly with the first form in this class. To see the proof of the Chain Rule see the Proof of Various Derivative Formulas section of the Extras chapter. Now, let’s go back and use the Chain Rule on the function that we used when we opened this section.

Example 1 Use the Chain Rule to differentiate ( ) 5 8R z z= − . Solution We’ve already identified the two functions that we needed for the composition, but let’s write them back down anyway and take their derivatives.

( ) ( )

( ) ( )

5 81 5

2

f z z g z z

f z g zz

= = −

′ ′= =

So, using the chain rule we get,

( ) ( )( ) ( )( ) ( )

( ) ( )

( )

12

5 81 5 8 52

1 52 5 8

52 5 8

R z f g z g z

f z g z

z

z

z

−

′ ′ ′=

′ ′= −

= −

=−

=−

And this is what we got using the definition of the derivative. In general we don’t really do all the composition stuff in using the Chain Rule. That can get a little complicated and in fact obscures the fact that there is a quick and easy way of remembering the chain rule that doesn’t require us to think in terms of function composition. Let’s take the function from the previous example and rewrite it slightly.

( ) ( ) 12

outside inside function function

5 8R z z= −14243

This function has an “inside function” and an “outside function”. The outside function is the square root or the exponent of 1

2 depending on how you want to think of it and the inside


Calculus I


function is the stuff that we’re taking the square root of or raising to the 12 , again depending on

how you want to look at it. The derivative is then,

( ) ( ) ( )

derivative ofoutside function

12

inside function derivative ofleft alone inside function

1 5 8 52

R z z −′ = −

6447448

14243 123

In general this is how we think of the chain rule. We identify the “inside function” and the “outside function”. We then differentiate the outside function leaving the inside function alone and multiply all of this by the derivative of the inside function. In its general form this is, ( ) ( )( ) ( )

derivative of times derivativeinside functionoutside function of inside functionleft alone

F x f g x g x′ ′ ′= 123 1424314243

We can always identify the “outside function” in the examples below by asking ourselves how we would evaluate the function. For instance in the R(z) case if we were to ask ourselves what R(2) is we would first evaluate the stuff under the radical and then finally take the square root of this result. The square root is the last operation that we perform in the evaluation and this is also the outside function. The outside function will always be the last operation you would perform if you were going to evaluate the function. Let’s take a look at some examples of the Chain Rule. Example 2 Differentiate each of the following.

(a) ( ) ( )2sin 3f x x x= + [Solution]

(b) ( ) ( )( )5032 cosf t t t= + [Solution]

(c) ( ) 4 23 9w wh w − += e [Solution]

(d) ( ) ( )4 4lng x x x−= + [Solution]

(e) ( )sec 1 5y x= − [Solution]

(f) ( ) ( ) ( )4 4cos cosP t t t= + [Solution]

Solution

(a) ( ) ( )2sin 3f x x x= +

It looks like the outside function is the sine and the inside function is 3x2+x. The derivative is then.

( ) ( ) ( )2

derivative of times derivativeleave insideoutside function of inside functionfunction alone

cos 3 6 1f x x x x′ = + +14243 1424314243


Calculus I


Or with a little rewriting,

( ) ( ) ( )26 1 cos 3f x x x x′ = + + [Return to Problems]

(b) ( ) ( )( )5032 cosf t t t= +

In this case the outside function is the exponent of 50 and the inside function is all the stuff on the inside of the parenthesis. The derivative is then.

( ) ( )( ) ( )( )

( )( ) ( )( )

493 2

492 3

50 2 cos 6 sin

50 6 sin 2 cos

f t t t t t

t t t t

′ = + −

= − +


(c) ( ) 4 23 9w wh w − += e

Identifying the outside function in the previous two was fairly simple since it really was the “outside” function in some sense. In this case we need to be a little careful. Recall that the outside function is the last operation that we would perform in an evaluation. In this case if we were to evaluate this function the last operation would be the exponential. Therefore the outside function is the exponential function and the inside function is its exponent. Here’s the derivative.

( ) ( )

( )

4 2

4 2

3 9 3

3 3 9

4 6

4 6

w w

w w

h w w w

w w

− +

− +

′ = −

= −

e

e

Remember, we leave the inside function alone when we differentiate the outside function. So, the derivative of the exponential function (with the inside left alone) is just the original function.


(d) ( ) ( )4 4lng x x x−= +

Here the outside function is the natural logarithm and the inside function is stuff on the inside of the logarithm.

( ) ( )5 3

5 34 4 4 4

1 4 44 4 x xg x x xx x x x

−−

− −

− +′ = − + =+ +

Again remember to leave the inside function along when differentiating the outside function. So, upon differentiating the logarithm we end up not with 1/x but instead with 1/(inside function).

[Return to Problems] (e) ( )sec 1 5y x= −

In this case the outside function is the secant and the inside is the 1 5x− .


Calculus I


( ) ( )( )

( ) ( )sec 1 5 tan 1 5 5

5sec 1 5 tan 1 5

y x x

x x

′ = − − −

= − − −

In this case the derivative of the outside function is ( ) ( )sec tanx x . However, since we leave the

inside function alone we don’t get x’s in both. Instead we get 1 5x− in both. [Return to Problems]

(f) ( ) ( ) ( )4 4cos cosP t t t= +

There are two points to this problem. First, there are two terms and each will require a different application of the chain rule. That will often be the case so don’t expect just a single chain rule when doing these problems. Second, we need to be very careful in choosing the outside and inside function for each term. Recall that the first term can actually be written as,

( ) ( )( )44cos cost t= So, in the first term the outside function is the exponent of 4 and the inside function is the cosine. In the second term it’s exactly the opposite. In the second term the outside function is the cosine and the inside function is 4t . Here’s the derivative for this function.

( ) ( ) ( )( ) ( )( )

( ) ( ) ( )

3 4 3

3 3 4

4cos sin sin 4

4sin cos 4 sin

P t t t t t

t t t t

′ = − −

= − −

[Return to Problems] There are a couple of general formulas that we can get for some special cases of the chain rule. Let’s take a quick look at those. Example 3 Differentiate each of the following.

(a) ( ) ( ) nf x g x=

(b) ( ) ( )g xf x = e

(c) ( ) ( )( )lnf x g x= Solution (a) The outside function is the exponent and the inside is g(x).

( ) ( ) ( )1nf x n g x g x

−′ ′=

(b) The outside function is the exponential function and the inside is g(x).

( ) ( ) ( )g xf x g x′ ′= e


Calculus I


(c) The outside function is the logarithm and the inside is g(x).

( ) ( ) ( ) ( )( )

1 g xf x g x

g x g x′

′ ′= =

The formulas in this example are really just special cases of the Chain Rule but may be useful to remember in order to quickly do some of these derivatives. Now, let’s also not forget the other rules that we’ve got for doing derivatives. For the most part we’ll not be explicitly identifying the inside and outside functions for the remainder of the problems in this section. We will be assuming that you can see our choices based on the previous examples and the work that we have shown. Example 4 Differentiate each of the following.

(a) ( ) ( ) 31 2tan 2 1 3T x x x−= − [Solution]

(b) ( )( )

53

32

4

1 2

xy

x

+=

− [Solution]

Solution

(a) ( ) ( ) 31 2tan 2 1 3T x x x−= −

This requires the product rule and each derivative in the product rule will require a chain rule application as well.

( )( )

( ) ( ) ( ) ( ) ( )

( )( )

( ) ( )

1 22 1 23 3

2

12 3 2

2 132

1 12 1 3 tan 2 1 3 631 2

2 1 32 1 3 tan 2

1 2

T x x x x xx

xx x x

x

−−

− −

′ = − + − − +

−= − −

+

In this part be careful with the inverse tangent. We know that,

( )12

1tan1

d xdx x

− =+

When doing the chain rule with this we remember that we’ve got to leave the inside function alone. That means that where we have the 2x in the derivative of 1tan x− we will need to have

( )2inside function . [Return to Problems]


Calculus I


(b) ( )( )

53

32

4

1 2

xy

x

+=

−

In this case we will be using the chain rule in concert with the quotient rule.

( ) ( )( ) ( ) ( ) ( ) ( )

( )( )4 3 5 23 2 2 3 2

232

5 4 3 1 2 4 3 1 2 4

1 2

x x x x x xy

x

+ − − + − −′ =

−

These tend to be a little messy. Notice that when we go to simplify that we’ll be able to a fair amount of factoring in the numerator and this will often greatly simplify the derivative.

( ) ( ) ( )( ) ( )( )( )( )( )

( ) ( )( )

4 23 2 2 2 3

62

43 3

42

4 1 2 5 3 1 2 4 3 4

1 2

3 4 5 6 16

1 2

x x x x x xy

x

x x x x

x

+ − − − + −′ =

−

+ − +=

−

After factoring we were able to cancel some of the terms in the numerator against the denominator. So even though the initial chain rule was fairly messy the final answer is significantly simpler because of the factoring.

[Return to Problems] The point of this last example is to not forget the other derivative rules that we’ve got. Most of the examples in this section won’t involve the product rule or the quotient rule to make the problems a little shorter. However, in practice they will often be in the same problem. Now, let’s take a look at some more complicated examples. Example 5 Differentiate each of the following.

(a) ( )( )109

2

4 zh z

z −=

+ e [Solution]

(b) ( ) ( )322 3 4f y y y y= + + [Solution]

(c) ( )( )3 2 4tan 3 ln 5y x x= + [Solution]

(d) ( ) ( )( )3 1sin 3sin 6tg t t−= +e [Solution]

Solution We’re going to be a little more careful in these problems than we were in the previous ones. The reason will be quickly apparent.


Calculus I


(a) ( )( )109

2

4 zh z

z −=

+ e

In this case let’s first rewrite the function in a form that will be a little easier to deal with.

( ) ( ) 1092 4 zh z z−−= + e

Now, let’s start the derivative.

( ) ( ) ( )119 920 4 4z zdh z z zdz

−− −′ = − + +e e

Notice that we didn’t actually do the derivative of the inside function yet. This is to allow us to notice that when we do differentiate the second term we will require the chain rule again. Notice as well that we will only need the chain rule on the exponential and not the first term. In many functions we will be using the chain rule more than once so don’t get excited about this when it happens. Let’s go ahead and finish this example out.

( ) ( ) ( )119 920 4 4 9z zh z z−− −′ = − + −e e

Be careful with the second application of the chain rule. Only the exponential gets multiplied by the “-9” since that’s the derivative of the inside function for that term only. One of the more common mistakes in these kinds of problems is to multiply the whole thing by the “-9” and not just the second term.


(b) ( ) ( )322 3 4f y y y y= + +

We’ll not put as many words into this example, but we’re still going to be careful with this derivative so make sure you can follow each of the steps here.

( ) ( )( ) ( )( )( )( ) ( ) ( )( )( )( ) ( )( )( )

13 322 2

13 222 2

13 222 2

1 2 3 4 2 3 42

1 2 3 4 2 3 3 4 3 82

1 2 3 4 2 9 24 3 42

df y y y y y y ydy

y y y y y y

y y y y y y

−

−

−

′ = + + + +

= + + + + +

= + + + + +

As with the first example the second term of the inside function required the chain rule to differentiate it. Also note that again we need to be careful when multiplying by the derivative of the inside function when doing the chain rule on the second term.



Calculus I


(c) ( )( )3 2 4tan 3 ln 5y x x= +

Let’s jump right into this one.

( )( ) ( ) ( )

( )( ) ( ) ( )

( ) ( )( )

132 2 4 2 43

2 332 2 4 2 3

4

232 2 2 43

sec 3 ln 5 3 ln 5

1 20sec 3 ln 5 3 63 5

42 3 sec 3 ln 5

dy x x x xdx

xx x x xx

x x x xx

−

−

′ = + +

= + +

= + +

In this example both of the terms in the inside function required a separate application of the chain rule.


(d) ( ) ( )( )3 1sin 3sin 6tg t t−= +e

We’ll need to be a little careful with this one.

( ) ( )( ) ( )( )

( )( ) ( )( ) ( )( )( )( ) ( )( ) ( ) ( )( )( )

( )( ) ( )( ) ( )( )

2 1 1

2 1 1 1

2 1 1 1

1 2 1 1

3sin 3sin 6 sin 3sin 6

3sin 3sin 6 cos 3sin 6 3sin 6

3sin 3sin 6 cos 3sin 6 1 3cos 6 6

3 18cos 6 sin 3sin 6 cos 3sin 6

t t

t t t

t t t

t t t

dg t t tdt

dt t tdt

t t t

t t t

− −

− − −

− − −

− − −

′ = + +

= + + +

= + + − +

= − + + +

e e

e e e

e e e

e e e

This problem required a total of 4 chain rules to complete.

[Return to Problems] Sometimes these can get quite unpleasant and require many applications of the chain rule. Initially, in these cases it’s usually best to be careful as we did in this previous set of examples and write out a couple of extra steps rather than trying to do it all in one step in your head. Once you get better at the chain rule you’ll find that you can do these fairly quickly in your head. Finally, before we move onto the next section there is one more issue that we need to address. In the Derivatives of Exponential and Logarithm Functions section we claimed that, ( ) ( ) ( )lnx xf x a f x a a′= ⇒ = but at the time we didn’t have the knowledge to do this. We now do. What we needed was the Chain Rule. First, notice that using a property of logarithms we can write a as, ln aa = e


Calculus I


This may seem kind of silly, but it is needed to compute the derivative. Now, using this we can write the function as,

( )( )

( )( )

ln

ln

ln

x

x

xa

a x

x a

f x a

a

=

=

=

=

=

e

ee

Okay, now that we’ve gotten that taken care of all we need to remember is that a is a constant and so ln a is also a constant. Now, differentiating the final version of this function is a (hopefully) fairly simple Chain Rule problem. ( ) ( )ln lnx af x a′ = e Now, all we need to do is rewrite the first term back as xa to get, ( ) ( )lnxf x a a′ = So, not too bad if you can see the trick to rewrite a and with the Chain Rule.


Calculus I


Implicit Differentiation To this point we’ve done quite a few derivatives, but they have all been derivatives of functions of the form ( )y f x= . Unfortunately not all the functions that we’re going to look at will fall

into this form. Let’s take a look at an example of a function like this. Example 1 Find y′ for 1xy = . Solution There are actually two solution methods for this problem. Solution 1 : This is the simple way of doing the problem. Just solve for y to get the function in the form that we’re used to dealing with and then differentiate.

2

1 1y yx x

′= ⇒ = −

So, that’s easy enough to do. However, there are some functions for which this can’t be done. That’s where the second solution technique comes into play. Solution 2 : In this case we’re going to leave the function in the form that we were given and work with it in that form. However, let’s recall from the first part of this solution that if we could solve for y then we will get y as a function of x. In other words, if we could solve for y (as we could in this case, but won’t always be able to do) we get ( )y y x= . Let’s rewrite the equation to note this.

( ) 1xy x y x= =

Be careful here and note that when we write ( )y x we don’t mean y times x. What we are noting

here is that y is some (probably unknown) function of x. This is important to recall when doing this solution technique. The next step in this solution is to differentiate both sides with respect to x as follows,

( )( ) ( )1d dx y xdx dx

=

The right side is easy. It’s just the derivative of a constant. The left side is also easy, but we’ve got to recognize that we’ve actually got a product here, the x and the ( )y x . So to do the

derivative of the left side we’ll need to do the product rule. Doing this gives,

( ) ( ) ( )( )1 0dy x x y xdx

+ =


Calculus I


Now, recall that we have the following notational way of writing the derivative.

( )( )d dyy x ydx dx

′= =

Using this we get the following, 0y xy′+ = Note that we dropped the ( )x on the y as it was only there to remind us that the y was a function

of x and now that we’ve taken the derivative it’s no longer really needed. We just wanted it in the equation to recognize the product rule when we took the derivative. So, let’s now recall just what were we after. We were after the derivative, y′ , and notice that there is now a y′ in the equation. So, to get the derivative all that we need to do is solve the equation for y′ .

yyx

′ = −

There it is. Using the second solution technique this is our answer. This is not what we got from the first solution however. Or at least it doesn’t look like the same derivative that we got from the first solution. Recall however, that we really do know what y is in terms of x and if we plug that in we will get,

2

1 1xyx x

′ = − = −

which is what we got from the first solution. Regardless of the solution technique used we should get the same derivative. The process that we used in the second solution to the previous example is called implicit differentiation and that is the subject of this section. In the previous example we were able to just solve for y and avoid implicit differentiation. However, in the remainder of the examples in this section we either won’t be able to solve for y or, as we’ll see in one of the examples below, the answer will not be in a form that we can deal with. In the second solution above we replaced the y with ( )y x and then did the derivative. Recall

that we did this to remind us that y is in fact a function of x. We’ll be doing this quite a bit in these problems, although we rarely actually write ( )y x . So, before we actually work anymore

implicit differentiation problems let’s do a quick set of “simple” derivatives that will hopefully help us with doing derivatives of functions that also contain a ( )y x .


Calculus I


Example 2 Differentiate each of the following.

(a) ( )535 7 1x x− + , ( ) 5f x , ( ) 5

y x [Solution]

(b) ( )sin 3 6x− , ( )( )sin y x [Solution]

(c) 2 9x x−e , ( )y xe [Solution]

Solution These are written a little differently from what we’re used to seeing here. This is because we want to match up these problems with what we’ll be doing in this section. Also, each of these parts has several functions to differentiate starting with a specific function followed by a general function. This again, is to help us with some specific parts of the implicit differentiation process that we’ll be doing.

(a) ( )535 7 1x x− + , ( ) 5f x , ( ) 5

y x

With the first function here we’re being asked to do the following,

( ) ( ) ( )5 43 3 25 7 1 5 5 7 1 15 7d x x x x xdx

− + = − + −

and this is just the chain rule. We differentiated the outside function (the exponent of 5) and then multiplied that by the derivative of the inside function (the stuff inside the parenthesis). For the second function we’re going to do basically the same thing. We’re going to need to use the chain rule. The outside function is still the exponent of 5 while the inside function this time is simply ( )f x . We don’t have a specific function here, but that doesn’t mean that we can’t at

least write down the chain rule for this function. Here is the derivative for this function,

( ) ( ) ( )5 45d f x f x f x

dx′=

We don’t actually know what ( )f x is so when we do the derivative of the inside function all we

can do is write down notation for the derivative, i.e. ( )f x′ .

With the final function here we simply replaced the f in the second function with a y since most of our work in this section will involve y’s instead of f’s. Outside of that this function is identical to the second. So, the derivative is,

( ) ( ) ( )5 45d y x y x y x

dx′=

[Return to Problems] (b) ( )sin 3 6x− , ( )( )sin y x

The first function to differentiate here is just a quick chain rule problem again so here is it’s derivative,


Calculus I


( ) ( )sin 3 6 6cos 3 6d x xdx

− = − −

For the second function we didn’t bother this time with using ( )f x and just jumped straight to

( )y x for the general version. This is still just a general version of what we did for the first

function. The outside function is still the sine and the inside is given by ( )y x and while we

don’t have a formula for ( )y x and so we can’t actually take its derivative we do have a notation

for its derivative. Here is the derivative for this function,

( )( ) ( ) ( )( )sin cosd y x y x y xdx

′=


(c) 2 9x x−e , ( )y xe

In this part we’ll just give the answers for each and leave out the explanation that we had in the first two parts.

( ) ( ) ( )( ) ( ) ( )2 29 92 9 y x y xx x x xd dx y xdx dx

− − ′= − =e e e e


So, in this set of examples we were just doing some chain rule problems where the inside function was ( )y x instead of a specific function. This kind of derivative shows up all the time in doing

implicit differentiation so we need to make sure that we can do them. Also note that we only did this for three kinds of functions but there are many more kinds of functions that we could have used here. So, it’s now time to do our first problem where implicit differentiation is required, unlike the first example where we could actually avoid implicit differentiation by solving for y. Example 3 Find y′ for the following function. 2 2 9x y+ = Solution Now, this is just a circle and we can solve for y which would give,

29y x= ± − Prior to starting this problem we stated that we had to do implicit differentiation here because we couldn’t just solve for y and yet that’s what we just did. So, why can’t we use “normal” differentiation here? The problem is the “ ± ”. With this in the “solution” for y we see that y is in fact two different functions. Which should we use? Should we use both? We only want a single function for the derivative and at best we have two functions here.


Calculus I


So, in this example we really are going to need to do implicit differentiation so we can avoid this. In this example we’ll do the same thing we did in the first example and remind ourselves that y is really a function of x and write y as ( )y x . Once we’ve done this all we need to do is

differentiate each term with respect to x.

( )( ) ( )22 9d dx y xdx dx

+ =

As with the first example the right side is easy. The left side is also pretty easy since all we need to do is take the derivative of each term and note that the second term will be similar the part (a) of the second example. All we need to do for the second term is use the chain rule. After taking the derivative we have,

( ) ( )12 2 0x y x y x′+ =

At this point we can drop the ( )x part as it was only in the problem to help with the

differentiation process. The final step is to simply solve the resulting equation for y′ .

2 2 0x yy

xyy

′+ =

′ = −

Unlike the first example we can’t just plug in for y since we wouldn’t know which of the two functions to use. Most answers from implicit differentiation will involve both x and y so don’t get excited about that when it happens. As always, we can’t forget our interpretations of derivatives. Example 4 Find the equation of the tangent line to 2 2 9x y+ =

at the point ( )2, 5 .

Solution First note that unlike all the other tangent line problems we’ve done in previous sections we need to be given both the x and the y values of the point. Notice as well that this point does lie on the graph of the circle (you can check by plugging the points into the equation) and so it’s okay to talk about the tangent line at this point. Recall that to write down the tangent line we need is the slope of the tangent line and this is nothing more than the derivative evaluated at the given point. We’ve got the derivative from the previous example so as we need to do is plug in the given point.


Calculus I


2, 5

25x y

m y= =

′= = −

The tangent line is then.

( )25 25

y x= − −

Now, let’s work some more examples. In the remaining examples we will no longer write ( )y x

for y. This is just something that we were doing to remind ourselves that y is really a function of x to help with the derivatives. Seeing the ( )y x reminded us that we needed to do the chain rule

on that portion of the problem. From this point on we’ll leave the y’s written as y’s and in our head we’ll need to remember that they really are ( )y x and that we’ll need to do the chain rule.

There is an easy way to remember how to do the chain rule in these problems. The chain rule really tells us to differentiate the function as we usually would, except we need to add on a derivative of the inside function. In implicit differentiation this means that every time we are differentiating a term with y in it the inside function is the y and we will need to add a y′ onto the term since that will be the derivative of the inside function. Let’s see a couple of examples. Example 5 Find y′ for each of the following.

(a) 3 5 33 8 1x y x y+ = + [Solution] (b) ( ) ( )2 10tan sec 2x y y x x+ = [Solution]

(c) ( )2 3 2 3lnx y x xy+ = −e [Solution]

Solution (a) 3 5 33 8 1x y x y+ = +

First differentiate both sides with respect to x and remember that each y is really ( )y x we just

aren’t going to write it that way anymore. This means that the first term on the left will be a product rule. We differentiated these kinds of functions involving y’s to a power with the chain rule in the Example 2 above. Also, recall the discussion prior to the start of this problem. When doing this kind of chain rule problem all that we need to do is differentiate the y’s as normal and then add on a y′ , which is nothing more than the derivative of the “inside function”. Here is the differentiation of each side for this function. 2 5 3 4 23 5 3 24x y x y y y y′ ′+ + =


Calculus I


Now all that we need to do is solve for the derivative, y′ . This is just basic solving algebra that you are capable of doing. The main problem is that it’s liable to be messier than what you’re used to doing. All we need to do is get all the terms with y′ in them on one side and all the terms without y′ in them on the other. Then factor y′ out of all the terms containing it and divide both sides by the “coefficient” of the y′ . Here is the solving work for this one,

( )2 5 2 3 4

2 5 2 3 4

2 5

2 3 4

3 3 24 5

3 3 24 5

3 324 5

x y y y x y y

x y y x y y

x yyy x y

′ ′+ = −

′+ = −

+′ =−

The algebra in these problems can be quite messy so be careful with that.

[Return to Problems] (b) ( ) ( )2 10tan sec 2x y y x x+ =

We’ve got two product rules to deal with this time. Here is the derivative of this function. ( ) ( ) ( ) ( ) ( )2 2 9 102 tan sec 10 sec sec tan 2x y x y y y y x y x x′ ′+ + + = Notice the derivative tacked onto the secant! Again, this is just a chain rule problem similar to the second part of Example 2 above. Now, solve for the derivative.

( ) ( )( ) ( ) ( ) ( )

( ) ( ) ( )( ) ( )

2 2 9 10

10

2 2 9

sec 10 sec 2 sec tan 2 tan

2 sec tan 2 tansec 10 sec

x y y x y y x x x y

y x x x yy

x y y x

′+ = − −

− −′ =

+

[Return to Problems] (c) ( )2 3 2 3lnx y x xy+ = −e

We’re going to need to be careful with this problem. We’ve got a couple chain rules that we’re going to need to deal with here that are a little different from those that we’ve dealt with prior to this problem. In both the exponential and the logarithm we’ve got a “standard” chain rule in that there is something other than just an x or y inside the exponential and logarithm. So, this means we’ll do the chain rule as usual here and then when we do the derivative of the inside function for each term we’ll have to deal with differentiating y’s. Here is the derivative of this equation.


Calculus I


( )3 2

2 33

32 3 2x y y xy yy xxy

+ ′+′+ = −e

In both of the chain rules note that the y′ didn’t get tacked on until we actually differentiated the y’s in that term. Now we need to solve for the derivative and this is liable to be somewhat messy. In order to get the y′ on one side we’ll need to multiply the exponential through the parenthesis and break up the quotient.

( )

3 22 3 2 3

3 3

2 3 2 3

2 3 1 1 2 3

1 2 3

2 3 1

32 3 2

1 32 3 2

3 3 2 2

2 23 3

x y x y

x y x y

x y x y

x y

x y

y xy yy xxy xy

yy xx y

y y x x

x xyy

+ +

+ +

+ − − +

− +

+ −

′′+ = − −

′′+ = − −

′+ = − −

− −′ =+

e e

e e

e e

ee

Note that to make the derivative at least look a little nicer we converted all the fractions to negative exponents.

[Return to Problems] Okay, we’ve seen one application of implicit differentiation in the tangent line example above. However, there is another application that we will be seeing in every problem in the next section. In some cases we will have two (or more) functions all of which are functions of a third variable. So, we might have ( )x t and ( )y t , for example and in these cases we will be differentiating

with respect to t. This is just implicit differentiation like we did in the previous examples, but there is a difference however. In the previous examples we have functions involving x’s and y’s and thinking of y as ( )y x . In

these problems we differentiated with respect to x and so when faced with x’s in the function we differentiated as normal and when faced with y’s we differentiated as normal except we then added a y′ onto that term because we were really doing a chain rule. In the new example we want to look at we’re assuming that ( )x x t= and that ( )y y t= and

differentiating with respect to t. This means that every time we are faced with an x or a y we’ll be doing the chain rule. This in turn means that when we differentiate an x we will need to add on an x′ and whenever we differentiate a y we will add on a y′ .


Calculus I


These new types of problems are really the same kind of problem we’ve been doing in this section. They are just expanded out a little to include more than one function that will require a chain rule. Let’s take a look at an example of this kind of problem. Example 6 Assume that ( )x x t= and ( )y y t= and differentiate the following equation with respect to t. ( )3 6 1 2cos 5xx y y y−+ − =e Solution So, just differentiate as normal and add on an appropriate derivative at each step. Note as well that the first term will be a product rule since both x and y are functions of t. ( )2 6 3 5 13 6 5 sin 5 2xx x y x y y x y y yy−′ ′ ′ ′ ′+ − + =e There really isn’t all that much to this problem. Since there are two derivatives in the problem we won’t be bothering to solve for one of them. When we do this kind of problem in the next section the problem will imply which one we need to solve for. At this point there doesn’t seem be any real reason for doing this kind of problem, but as we’ll see in the next section every problem that we’ll be doing there will involve this kind of implicit differentiation.


Calculus I


Related Rates In this section we are going to look at an application of implicit differentiation. Most of the applications of derivatives are in the next chapter however there are a couple of reasons for placing it in this chapter as opposed to putting it into the next chapter with the other applications. The first reason is that it’s an application of implicit differentiation and so putting it right after that section means that we won’t have forgotten how to do implicit differentiation. The other reason is simply that after doing all these derivatives we need to be reminded that there really are actual applications to derivatives. Sometimes it is easy to forget there really is a reason that we’re spending all this time on derivatives. For these related rates problems it’s usually best to just jump right into some problems and see how they work. Example 1 Air is being pumped into a spherical balloon at a rate of 5 cm3/min. Determine the rate at which the radius of the balloon is increasing when the diameter of the balloon is 20 cm. Solution The first thing that we’ll need to do here is to identify what information that we’ve been given and what we want to find. Before we do that let’s notice that both the volume of the balloon and the radius of the balloon will vary with time and so are really functions of time, i.e. ( )V t and

( )r t .

We know that air is being pumped into the balloon at a rate of 5 cm3/min. This is the rate at which the volume is increasing. Recall that rates of change are nothing more than derivatives and so we know that, ( ) 5V t′ = We want to determine the rate at which the radius is changing. Again, rates are derivatives and so it looks like we want to determine,

( ) ( )? when 10cm2dr t r t′ = = =

Note that we needed to convert the diameter to a radius. Now that we’ve identified what we have been given and what we want to find we need to relate these two quantities to each other. In this case we can relate the volume and the radius with the formula for the volume of a sphere.

( ) ( ) 343

V t r tπ=

As in the previous section when we looked at implicit differentiation, we will typically not use the ( )t part of things in the formulas, but since this is the first time through one of these we will


Calculus I


do that to remind ourselves that they are really functions of t. Now we don’t really want a relationship between the volume and the radius. What we really want is a relationship between their derivatives. We can do this by differentiating both sides with respect to t. In other words, we will need to do implicit differentiation on the above formula. Doing this gives, 24V r rπ′ ′= Note that at this point we went ahead and dropped the ( )t from each of the terms. Now all that

we need to do is plug in what we know and solve for what we want to find.

( )2 15 4 10 cm/min80

r rππ

′ ′= ⇒ =

We can get the units of the derivative by recalling that,

drrdt

′ =

The units of the derivative will be the units of the numerator (cm in the previous example) divided by the units of the denominator (min in the previous example). Let’s work some more examples. Example 2 A 15 foot ladder is resting against the wall. The bottom is initially 10 feet away from the wall and is being pushed towards the wall at a rate of 1

4 ft/sec. How fast is the top of the ladder moving up the wall 12 seconds after we start pushing? Solution The first thing to do in this case is to sketch picture that shows us what is going on.

We’ve defined the distance of the bottom of the latter from the wall to be x and the distance of the top of the ladder from the floor to be y. Note as well that these are changing with time and so we really should write ( )x t and ( )y t . However, as is often the case with related rates/implicit

differentiation problems we don’t write the ( )t part just try to remember this in our heads as we

proceed with the problem. Next we need to identify what we know and what we want to find. We know that the rate at which the bottom of the ladder is moving towards the wall. This is,


Calculus I


14

x′ = −

Note as well that the rate is negative since the distance from the wall, x, is decreasing. We always need to be careful with signs with these problems. We want to find the rate at which the top of the ladder is moving away from the floor. This is y′ . Note as well that this quantity should be positive since y will be increasing. As with the first example we first need a relationship between x and y. We can get this using Pythagorean theorem.

( )22 2 15 225x y+ = = All that we need to do at this point is to differentiate both sides with respect to t, remembering that x and y are really functions of t and so we’ll need to do implicit differentiation. Doing this gives an equation that shows the relationship between the derivatives. 2 2 0xx yy′ ′+ = (1) Next, let’s see which of the various parts of this equation that we know and what we need to find. We know x′ and are being asked to determine y′ so it’s okay that we don’t know that. However, we still need to determine x and y. Determining x and y is actually fairly simple. We know that initially 10x = and the end is being pushed in towards the wall at a rate of 1

4 ft/sec and that we are interested in what has happened after 12 seconds. We know that,

( )

distance rate time1 12 34

= ×

= =

So, the end of the ladder has been pushed in 3 feet and so after 12 seconds we must have 7x = . Note that we could have computed this in one step as follows,

( )110 12 74

x = − =

To find y (after 12 seconds) all that we need to do is reuse the Pythagorean Theorem with the values of x that we just found above.

2225 225 49 176y x= − = − = Now all that we need to do is plug into (1) and solve for y′ .


Calculus I


( ) ( )71 742 7 2 176 0 0.1319 ft/sec

4 176 4 176y y ′ ′− + = ⇒ = = =

Notice that we got the correct sign for y′ . If we’d gotten a negative then we’d have known that we had made a mistake and we could go back and look for it. Example 3 Two people are 50 feet apart. One of them starts walking north at a rate so that the angle shown in the diagram below is changing at a constant rate of 0.01 rad/min. At what rate is distance between the two people changing when 0.5θ = radians?

Solution This example is not as tricky as it might at first appear. Let’s call the distance between them at any point in time x as noted above. We can then relate all the known quantities by one of two trig formulas.

50cos sec50x

xθ θ= =

We want to find x′ and we could find x if we wanted to at the point in question using cosine since we also know the angle at that point in time. However, if we use the second formula we won’t need to know x as you’ll see. So, let’s differentiate that formula.

sec tan50x

θ θ θ′

′ =

As noted, there are no x’s in this formula. We want to determine x′ and we know that 0.5θ = and 0.01θ ′ = (do you agree with it being positive?). So, just plug in and solve. ( )( ) ( ) ( )50 0.01 sec 0.5 tan 0.5 0.311254 ft / minx x′ ′= ⇒ = So far we we’ve seen three related rates problems. While each one was worked in a very different manner the process was essentially the same in each. In each problem we identified what we were given and what we wanted to find. We next wrote down a relationship between all the various quantities and used implicit differentiation to arrive at a relationship between the various derivatives in the problem. Finally, we plugged into the equation to find the value we were after. So, in a general sense each problem was worked in pretty much the same manner. The only real difference between them was coming up with the relationship between the known and unknown


Calculus I


quantities. This is often the hardest part of the problem. In many problems the best way to come up with the relationship is to sketch a diagram that shows the situation. This often seems like a silly step, but can make all the difference in whether we can find the relationship or not. Let’s work another problem that uses some different ideas and shows some of the different kinds of things that can show up in related rates problems. Example 4 A tank of water in the shape of a cone is leaking water at a constant rate of

32ft /hour . The base radius of the tank is 5 ft and the height of the tank is 14 ft. (a) At what rate is the depth of the water in the tank changing when the depth of the

water is 6 ft?

(b) At what rate is the radius of the top of the water in the tank changing when the depth of the water is 6 ft?

Solution Okay, we should probably start off with a quick sketch (probably not to scale) of what is going on here.

As we can see, the water in the tank actually forms a smaller cone with the same central angle as the tank itself. The radius of the “water” cone at any time is given by r and the height of the “water” cone at any time is given by h. The volume of water in the tank at any time t is given by,

213

V r hπ=

and we’ve been given that 2V ′ = − . (a) At what rate is the depth of the water in the tank changing when the depth of the water is 6 ft? For this part we need to determine h′ when 6h = and now we have a problem. The only


Calculus I


formula that we’ve got that will relate the volume to the height also includes the radius and so if we were to differentiate this with respect to t we would get,

22 13 3

V rr h r hπ π′ ′ ′= +

So, in this equation we know V ′ and h and want to find h′ , but we don’t know r and r′ . As we’ll see finding r isn’t too bad, but we just don’t have enough information, at this point, that will allow us to find r′ and h′ simultaneously. To fix this we’ll need to eliminate the r from the volume formula in some way. This is actually easier than it might at first look. If we go back to our sketch above and look at just the right half of the tank we see that we have two similar triangles and when we say similar we mean similar in the geometric sense. Recall that two triangles are called similar if their angles are identical, which is the case here. When we have two similar triangles then ratios of any two sides will be equal. For our set this means that we have,

5 514 14

r r hh

= ⇒ =

If we take this and plug it into our volume formula we have,

2

2 31 1 5 253 3 14 588

V r h h h hπ π π = = =

This gives us a volume formula that only involved the volume and the height of the water. Note however that this volume formula is only valid for our cone, so don’t be tempted to use it for other cones! If we now differentiate this we have,

225196

V h hπ′ ′=

At this point all we need to do is plug in what we know and solve for h′ .

( )225 982 6 0.1386196 225

h hππ

−′ ′− = ⇒ = = −

So, it looks like the height is decreasing at a rate of 0.1386 ft/hr. (b) At what rate is the radius of the top of the water in the tank changing when the depth of the water is 6 ft? In this case we are asking for r′ and there is an easy way to do this part and a difficult (well, more difficult than the easy way anyway….) way to do it. The “difficult” way is to redo the work in part (a) above only this time use,

14 145 5

h h rr

= ⇒ =


Calculus I


to get the volume in terms of V and r and then proceed as before. That’s not terribly difficult, but it is more work that we need to so. Recall from the first part that we have,

5 5

14 14r h r h′ ′= ⇒ =

So, as we can see if we take the relationship that relates r and h that we used in the first part and differentiate it we get a relationship between r′ and h′ . At this point all we need to do here is use the result from the first part to get,

5 98 7 0.04951

14 225 45r

π π− ′ = = − = −

Much easier that redoing all of the first part. Note however, that we were only able to do this the “easier” way because it was asking for r′ at exactly the same time that we asked for h′ in the first part. If we hadn’t been using the same time then we would have had no choice but to do this the “difficult” way. In the second part of the previous problem we saw an important idea in dealing with related rates. In order to find the asked for rate all we need is an equation that relates the rate we’re looking for to a rate that we already know. Sometimes there are multiple equations that we can use and sometimes one will be easier than another. Also, this problem showed us that we will often have an equation that contains more variables that we have information about and so, in these cases, we will need to eliminate one (or more) of the variables. In this problem we eliminated the extra variable using the idea of similar triangles. This will not always be how we do this, but many of these problems do use similar triangles so make sure you can use that idea. Let’s work some more problems. Example 5 A trough of water is 8 meters deep and its ends are in the shape of isosceles triangles whose width is 5 meters and height is 2 meters. If water is being pumped in at a constant rate of 36 m /sec . At what rate is the height of the water changing when the water has a height of 120 cm? Solution Note that an isosceles triangle is just a triangle in which two of the sides are the same length. In our case sides of the tank have the same length. We definitely need a sketch of this situation to get us going here so here. A sketch of the trough is shown below.


Calculus I


Now, in this problem we know that 36 m /secV ′ = and we want to determine h′ when

1.2 mh = . Note that because V ′ is in terms of meters we need to convert h into meters as well. So, we need an equation that will relate these two quantities and the volume of the tank will do it. The volume of this kind of tank is simple to compute. The volume is the area of the end times the depth. For our case the volume of the water in the tank is,

( ) ( )( )( )

( )

12

12

Area of End depth

base height depth

84

V

hwhw

=

= ×

=

=

As with the previous example we’ve got an extra quantity here, w, that is also changing with time and so we need to eliminate it from the problem. To do this we’ll again make use of the idea of similar triangles. If we look at the end of the tank we’ll see that we again have two similar triangles. One for the tank itself and one formed by the water in the tank. Again, remember that with similar triangles ratios of sides must be equal. In our case we’ll use,

55 2 2w h w h= ⇒ =

Plugging this into the volume gives a formula for the volume (and only for this tank) that only involved the height of the water.

254 4 102

V hw h h h = = =

We can now differentiate this to get, 20V hh′ ′= Finally, all we need to do is plug in and solve for h′ .


Calculus I


( )6 20 1.2 0.25 m/sech h′ ′= ⇒ =

So, the height of the water is rising at a rate of 0.25 m/sec. Example 6 A light is on the top of a 12 ft tall pole and a 5ft 6in tall person is walking away from the pole at a rate of 2 ft/sec.

(a) At what rate is the tip of the shadow moving away from the pole when the person is 25 ft from the pole?

(b) At what rate is the tip of the shadow moving away from the person when the person is 25 ft from the pole?

Solution We’ll definitely need a sketch of this situation to get us started here. The tip of the shadow is defined by the rays of light just getting past the person and so we can form the following set of similar triangles.

Here x is the distance of the tip of the shadow from the pole, px is the distance of the person

from the pole and sx is the length of the shadow. Also note that we converted the persons height over to 5.5 feet since all the other measurements are in feet. (a) At what rate is the tip of the shadow moving away from the pole when the person is 25 ft from the pole? In this case we want to determine x′ when 25px = given that 2px′ = .

The equation we’ll need here is, p sx x x= +

but we’ll need to eliminate sx from the equation in order to get an answer. To do this we can again make use of the fact that the two triangles are similar to get,


Calculus I


1125.5 5.5 11Note :

12 12 12 24p

s s

s

x xx x x

= = = =+

We’ll need to solve this for sx .

( )1124

11 1324 241113

p

p

p

s s

s

s

x x x

x x

x x

+ =

=

=

Our equation then becomes,

11 2413 13p p px x x x= + =

Now all that we need to do is differentiate this, plug in and solve for x′ .

( )24 24 2 3.6923 ft/sec13 13px x x′ ′ ′= ⇒ = =

The tip of the shadow is then moving away from the pole at a rate of 3.6923 ft/sec. Notice as well that we never actually had to use the fact that 25px = for this problem. That will happen

on rare occasions. (b) At what rate is the tip of the shadow moving away from the person when the person is 25 ft from the pole? This part is actually quite simple if we have the answer from (a) in hand, which we do of course. In this case we know that sx represents the length of the shadow, or the distance of the tip of the

shadow from the person so it looks like we want to determine sx′ when 25px = .

Again, we can use p sx x x= + , however unlike the first part we now know that 2px′ = and

3.6923 ft/secx′ = so in this case all we need to do is differentiate the equation and plug in for all the known quantities.

3.6923 2 1.6923 ft/sec

p s

s s

x x x

x x

′ ′ ′= +

′ ′= + =

The tip of the shadow is then moving away from the person at a rate of 1.6923 ft/sec.


Calculus I


Example 7 A spot light is on the ground 20 ft away from a wall and a 6 ft tall person is walking towards the wall at a rate of 2.5 ft/sec. How fast is the height of the shadow changing when the person is 8 feet from the wall? Is the shadow increasing or decreasing in height at this time? Solution Let’s start off with a sketch of this situation and the sketch here will be similar to that of the previous problem. The top of the shadow will be defined by the light rays going over the head of the person and so we again get yet another set of similar triangles.

In this case we want to determine y′ when the person is 8 ft from wall or 12 ftx = . Also, if the person is moving towards the wall at 2.5 ft/sec then the person must be moving away from the spotlight at 2.5 ft/sec and so we also know that 2.5x′ = . In all the previous problems that used similar triangles we used the similar triangles to eliminate one of the variables from the equation we were working with. In this case however, we can get the equation that relates x and y directly from the two similar triangles. In this case the equation we’re going to work with is,

20 1206y y

x x= ⇒ =

Now all that we need to do is differentiate and plug values into solve to get y′ .

( )2 2

120 120 2.5 2.0833 ft/sec12

y x yx

′ ′ ′= − ⇒ = − = −

The height of the shadow is then decreasing at a rate of 2.0833 ft/sec. Okay, we’ve worked quite a few problems now that involved similar triangles in one form or another so make sure you can do these kinds of problems. It’s now time to do a problem that while similar to some of the problems we’ve done to this point is also sufficiently different that it can cause problems until you’ve seen how to do it.


Calculus I


Example 8 Two people on bikes are separated by 350 meters. Person A starts riding north at a rate of 5 m/sec and 7 minutes later Person B starts riding south at 3 m/sec. At what rate is the distance separating the two people changing 25 minutes after Person A starts riding? Solution There is a lot to digest here with this problem. Let’s start off with a sketch of the situation.

Now we are after z′ and we know that 5x′ = and 3y′ = . We want to know z′ after Person A had been riding for 25 minutes and Person B has been riding for 25 7 18− = minutes. After converting these times to seconds (because our rates are all in m/sec) this means that at the time we’re interested in each of the bike riders has rode, ( ) ( )5 25 60 7500 m 3 18 60 3240 mx y= × = = × =

Next, the Pythagorean theorem tells us that,

( )22 2350z x y= + + (2)

Therefore, 25 minutes after Person A starts riding the two bike riders are

( ) ( )2 22 2350 7500 3240 350 10745.7015 mz x y= + + = + + =

apart. To determine the rate at which the two riders are moving apart all we need to do then is differentiate (2) and plug in all the quantities that we know to find z′ .


Calculus I


( )( )( ) ( )( )

2 2

2 10745.7015 2 7500 3240 5 37.9958 m/sec

zz x y x y

zz

′ ′ ′= + +

′ = + +

′ =

So, the two riders are moving apart at a rate of 7.9958 m/sec. Every problem that we’ve worked to this point has come down to needing a geometric formula and we should probably work a quick problem that is not geometric in nature. Example 9 Suppose that we have two resistors connected in parallel with resistances 1R and

2R measured in ohms ( Ω ). The total resistance, R, is then given by,

1 2

1 1 1R R R

= +

Suppose that 1R is increasing at a rate of 0.4 Ω /min and 2R is decreasing at a rate of

0.7 Ω /min. At what rate is R changing when 1 80R = Ω and 2 105R = Ω ?

Solution Okay, unlike the previous problems there really isn’t a whole lot to do here. First, let’s note that we’re looking for R′ and that we know 1 0.4R′ = and 2 0.7R′ = − . Be careful with the signs

here. Also, since we’ll eventually need it let’s determine R at the time we’re interested in.

1 1 1 37 1680 45.4054

80 105 1680 37R

R= + = ⇒ = = Ω

Next we need to differentiate the equation given in the problem statement.

( ) ( )

( ) ( )

1 22 22

1 2

21 22 2

1 2

1 1 1

1 1

R R RR R R

R R R RR R

′ ′ ′− = − −

′ ′ ′= +

Finally, all we need to do is plug into this and do some quick computations.

( ) ( ) ( )22 2

1 145.4054 0.4 0.7 0.00204580 105

R ′ = + − = −

So, it looks like R is decreasing at a rate of 0.002045 Ω /min.


Calculus I


We’ve seen quite a few related rates problems in this section that cover a wide variety of possible problems. There are still many more different kinds of related rates problems out there in the world, but the ones that we’ve worked here should give you a pretty good idea on how to at least start most of the problems that you’re liable to run into.


Calculus I


Higher Order Derivatives Let’s start this section with the following function. ( ) 3 25 3 10 5f x x x x= − + − By this point we should be able to differentiate this function without any problems. Doing this we get, ( ) 215 6 10f x x x′ = − + Now, this is a function and so it can be differentiated. Here is the notation that we’ll use for that, as well as the derivative.

( ) ( )( ) 30 6f x f x x′′′ ′= = −

This is called the second derivative and ( )f x′ is now called the first derivative.

Again, this is a function so we can differentiate it again. This will be called the third derivative. Here is that derivative as well as the notation for the third derivative.

( ) ( )( ) 30f x f x ′′′′ ′′= = Continuing, we can differentiate again. This is called, oddly enough, the fourth derivative. We’re also going to be changing notation at this point. We can keep adding on primes, but that will get cumbersome after awhile.

( ) ( )( )(4) 0f x f x ′′′′= = This process can continue but notice that we will get zero for all derivatives after this point. This set of derivatives leads us to the following fact about the differentiation of polynomials. Fact If p(x) is a polynomial of degree n (i.e. the largest exponent in the polynomial) then,

( ) ( ) 0 for 1kp x k n= ≥ + We will need to be careful with the “non-prime” notation for derivatives. Consider each of the following.

( ) ( ) ( )

( ) ( )

2

22

f x f x

f x f x

′′=

=

The presence of parenthesis in the exponent denotes differentiation while the absence of parenthesis denotes exponentiation. Collectively the second, third, fourth, etc. derivatives are called higher order derivatives.


Calculus I


Let’s take a look at some examples of higher order derivatives. Example 1 Find the first four derivatives for each of the following.

(a) ( )1

2 23 8 tR t t t= + + e [Solution] (b) cosy x= [Solution] (c) ( ) ( ) ( )2sin 3 ln 7yf y y y−= + +e [Solution]

Solution

(a) ( )1

2 23 8 tR t t t= + + e

There really isn’t a lot to do here other than do the derivatives.

( )

( )

( )

( )

12

32

52

7(4) 2

6 4

6 2

3

152

t

t

t

t

R t t t

R t t

R t t

R t t

−

−

−

−

′ = + +

′′ = − +

′′′ = +

= − +

e

e

e

e

Notice that differentiating an exponential function is very simple. It doesn’t change with each differentiation.

[Return to Problems] (b) cosy x= Again, let’s just do some derivatives.

( )4

cossincos

sin

cos

y xy xy xy x

y x

=′ = −′′ = −′′′ =

=

Note that cosine (and sine) will repeat every four derivatives. The other four trig functions will not exhibit this behavior. You might want to take a few derivatives to convince yourself of this.

[Return to Problems] (c) ( ) ( ) ( )2sin 3 ln 7yf y y y−= + +e

In the previous two examples we saw some patterns in the differentiation of exponential functions, cosines and sines. We need to be careful however since they only work if there is just a t or an x in the argument. This is the point of this example. In this example we will need to use the chain rule on each derivative.


Calculus I


( ) ( ) ( )

( ) ( )( ) ( )

( ) ( ) ( )

2 2 1

2 2

2 3

4 2 4

13cos 3 2 3cos 3 2

9sin 3 4

27cos 3 8 2

81sin 3 16 6

y y

y

y

y

f y y y yy

f y y y

f y y y

f y y y

− − −

− −

− −

− −

′ = − + = − +

′′ = − + −

′′′ = − − +

= + −

e e

e

e

e

So, we can see with slightly more complicated arguments the patterns that we saw for exponential functions, sines and cosines no longer completely hold.

[Return to Problems] Let’s do a couple more examples to make a couple of points. Example 2 Find the second derivative for each of the following functions.

(a) ( ) ( )sec 5Q t t= [Solution]

(b) ( ) 31 2wg w −= e [Solution]

(c) ( ) ( )2ln 1f t t= + [Solution]

Solution (a) ( ) ( )sec 5Q t t=

Here’s the first derivative. ( ) ( ) ( )5sec 5 tan 5Q t t t′ = Notice that the second derivative will now require the product rule.

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

2

2 3

25sec 5 tan 5 tan 5 25sec 5 sec 5

25sec 5 tan 5 25sec 5

Q t t t t t t

t t t

′′ = +

= +

Notice that each successive derivative will require a product and/or chain rule and that as noted above this will not end up returning back to just a secant after four (or another other number for that matter) derivatives as sine and cosine will.


(b) ( ) 31 2wg w −= e

Again, let’s start with the first derivative.

( ) 32 1 26 wg w w −′ = − e As with the first example we will need the product rule for the second derivative.

( ) ( )3 3

3 3

2 2

4

1 2 1 2

1 2 1 2

12 6 6

12 36

w w

w w

g w w w w

w w

− −

− −

′′ = − − −

= − +

e e

e e



Calculus I


(c) ( ) ( )2ln 1f t t= +

Same thing here.

( ) 2

21

tf tt

′ =+

The second derivative this time will require the quotient rule.

( ) ( ) ( ) ( )( )

( )

2

22

2

22

2 1 2 2

1

2 2

1

t t tf t

t

t

t

+ −′′ =

+

−=

+

[Return to Problems] As we saw in this last set of examples we will often need to use the product or quotient rule for the higher order derivatives, even when the first derivative didn’t require these rules. Let’s work one more example that will illustrate how to use implicit differentiation to find higher order derivatives. Example 3 Find y′′ for 2 4 10x y+ = Solution Okay, we know that in order to get the second derivative we need the first derivative and in order to get that we’ll need to do implicit differentiation. Here is the work for that.

3

3

2 4 0

2

x y yxyy

′+ =

′ = −

Now, this is the first derivative. We get the second derivative by differentiating this, which will require implicit differentiation again.

( )( )

3

3 2

23

3 2

6

4

2

2 6

2

2 64

32

xyy

y x y y

y

y xy yy

y xyy

′ ′′ = −

′−= −

′−= −

′−= −


Calculus I


This is fine as far as it goes. However, we would like there to be no derivatives in the answer. We don’t, generally, mind having x’s and/or y’s in the answer when doing implicit differentiation, but we really don’t like derivatives in the answer. We can get rid of the derivative however by acknowledging that we know what the first derivative is and substituting this into the second derivative equation. Doing this gives,

4

3

4

2 3

4

32

32

2322

y xyyy

xy xy

y

y x y

y

−

′−′′ = −

− −

= −

+= −

Now that we’ve found some higher order derivatives we should probably talk about an interpretation of the second derivative. If the position of an object is given by s(t) we know that the velocity is the first derivative of the position. ( ) ( )v t s t′= The acceleration of the object is the first derivative of the velocity, but since this is the first derivative of the position function we can also think of the acceleration as the second derivative of the position function. ( ) ( ) ( )a t v t s t′ ′′= =

Alternate Notation There is some alternate notation for higher order derivatives as well. Recall that there was a fractional notation for the first derivative.

( ) dff xdx

′ =

We can extend this to higher order derivatives.

( ) ( )2 3

2 3 .d y d yf x f x etcdx dx

′′ ′′′= =


Calculus I


Logarithmic Differentiation There is one last topic to discuss in this section. Taking the derivatives of some complicated functions can be simplified by using logarithms. This is called logarithmic differentiation. It’s easiest to see how this works in an example. Example 1 Differentiate the function.

( )

5

21 10 2xy

x x=

− +

Solution Differentiating this function could be done with a product rule and a quotient rule. However, that would be a fairly messy process. We can simplify things somewhat by taking logarithms of both sides.

( )

5

2ln ln

1 10 2xy

x x

= − +

Of course, this isn’t really simpler. What we need to do is use the properties of logarithms to expand the right side as follows.

( ) ( )( )( ) ( ) ( )

5 2

5 2

ln ln ln 1 10 2

ln ln ln 1 10 ln 2

y x x x

y x x x

= − − +

= − − − +

This doesn’t look all that simple. However, the differentiation process will be simpler. What we need to do at this point is differentiate both sides with respect to x. Note that this is really implicit differentiation.

( ) ( )

( )

12 24

152 2

2

1 2 25 10 21 10 2

5 101 10 2

x xy xy x x x

y xy x x x

−+′ −

= − −−

+

′= + −

− +

To finish the problem all that we need to do is multiply both sides by y and the plug in for y since we do know what that is.

( )

2

5

22

5 101 10 2

5 101 10 21 10 2

xy yx x x

x xx x xx x

′ = + − − +

= + − − + − +


Calculus I


Depending upon the person, doing this would probably be slightly easier than doing both the product and quotient rule. The answer is almost definitely simpler than what we would have gotten using the product and quotient rule. So, as the first example has shown we can use logarithmic differentiation to avoid using the product rule and/or quotient rule. We can also use logarithmic differentiation to differentiation functions in the form.

( )( ) ( )g xy f x=

Let’s take a quick look at a simple example of this. Example 2 Differentiate xy x= Solution We’ve seen two functions similar to this at this point.

( ) ( )1 lnn n x xd dx nx a a adx dx

−= =

Neither of these two will work here since both require either the base or the exponent to be a constant. In this case both the base and the exponent are variables and so we have no way to differentiate this function using only known rules from previous sections. With logarithmic differentiation we can do this however. First take the logarithm of both sides as we did in the first example and use the logarithm properties to simplify things a little.

ln lnln ln

xy xy x x

==

Differentiate both sides using implicit differentiation.

1ln ln 1y x x xy x′ = + = +

As with the first example multiply by y and substitute back in for y.

( )( )1 ln

1 lnx

y y x

x x

′ = +

= +

Let’s take a look at a more complicated example of this.


Calculus I


Example 3 Differentiate ( ) ( )cos1 3 xy x= − Solution Now, this looks much more complicated than the previous example, but is in fact only slightly more complicated. The process is pretty much identical so we first take the log of both sides and then simplify the right side. ( ) ( ) ( ) ( )cosln ln 1 3 cos ln 1 3xy x x x = − = −

Next, do some implicit differentiation.

( ) ( ) ( ) ( ) ( ) ( )3 3sin ln 1 3 cos sin ln 1 3 cos1 3 1 3

y x x x x x xy x x′ −

= − − + = − − −− −

Finally, solve for y′ and substitute back in for y.

( ) ( ) ( )

( ) ( ) ( ) ( ) ( )cos

3sin ln 1 3 cos1 3

31 3 sin ln 1 3 cos1 3

x

y y x x xx

x x x xx

′ = − − + − = − − − + −

A messy answer but there it is. We’ll close this section out with a quick recap of all the various ways we’ve seen of differentiating functions with exponents. It is important to not get all of these confused.

( )

( )

( )

( ) ( )

1

0 This is a constant

Power Rule

ln Derivative of an exponential function

1 ln Logarithmic Differentiation

b

n n

x x

x x

d adxd x nxdxd a a adxd x x xdx

−

=

=

=

= +

It is sometimes easy to get these various functions confused and use the wrong rule for differentiation. Always remember that each rule has very specific rules for where the variable and constants must be. For example, the Power Rule requires that the base be a variable and the exponent be a constant, while the exponential function requires exactly the opposite. If you can keep straight all the rules you can’t go wrong with these.


Calculus I



CALCULUS I Practice Problems

Derivatives

Paul Dawkins

Calculus I


Table of Contents Preface ............................................................................................................................................. 1 Derivatives ...................................................................................................................................... 2

Introduction .............................................................................................................................................. 2 The Definition of the Derivative .............................................................................................................. 3 Interpretations of the Derivative ............................................................................................................. 4 Differentiation Formulas .......................................................................................................................... 6 Product and Quotient Rule ....................................................................................................................... 8 Derivatives of Trig Functions .................................................................................................................. 9 Derivatives of Exponential and Logarithm Functions ......................................................................... 10 Derivatives of Inverse Trig Functions ................................................................................................... 10 Derivatives of Hyperbolic Functions ..................................................................................................... 11 Chain Rule ............................................................................................................................................... 11 Implicit Differentiation........................................................................................................................... 13 Related Rates .......................................................................................................................................... 14 Higher Order Derivatives ....................................................................................................................... 16 Logarithmic Differentiation ................................................................................................................... 17

Preface Here are a set of practice problems for my Calculus I notes. If you are viewing the pdf version of this document (as opposed to viewing it on the web) this document contains only the problems themselves and no solutions are included in this document. Solutions can be found in a number of places on the site.

1. If you’d like a pdf document containing the solutions go to the note page for the section you’d like solutions for and select the download solutions link from there. Or,

2. Go to the download page for the site http://tutorial.math.lamar.edu/download.aspx and select the section you’d like solutions for and a link will be provided there.

3. If you’d like to view the solutions on the web or solutions to an individual problem you can go to the problem set web page, select the problem you want the solution for. At this point I do not provide pdf versions of individual solutions, but for a particular problem you can select “Printable View” from the “Solution Pane Options” to get a printable version.

Note that some sections will have more problems than others and some will have more or less of a variety of problems. Most sections should have a range of difficulty levels in the problems although this will vary from section to section. I add problems to this document as I get a chance and so that means that sometimes some sections are apparently given more attention than others. Eventually I’ll have a full complement of problems for all the sections, however it takes time to write the problems, write the solutions for


http://tutorial.math.lamar.edu/download.aspx

Calculus I


each problem and format them for presentation on the web. If there aren’t practice problems for a given section (or there aren’t that many problems in a section) then know that I will eventually get around to writing problems for the section (or more problems for the section as the case may be). Please do not email me asking when I’m going to get problems for a particular section written. As I’ve already said, I write problems as I have time to do so. Please understand that I have my own classes to teach and other responsibilities to my department that I have to also take care of. This web site and all of its information has been written and maintained in my spare time and that often means that I don’t have as much time to work on it as I’d like to.

Derivatives

Introduction Here are a set of practice problems for the Derivatives chapter of my Calculus I notes. If you are viewing the pdf version of this document (as opposed to viewing it on the web) this document contains only the problems themselves and no solutions are included in this document. Solutions can be found in a number of places on the site.




Note that some sections will have more problems than others and some will have more or less of a variety of problems. Most sections should have a range of difficulty levels in the problems although this will vary from section to section. Here is a list of topics in this chapter that have practice problems written for them. The Definition of the Derivative Interpretation of the Derivative Differentiation Formulas Product and Quotient Rule Derivatives of Trig Functions



Calculus I


Derivatives of Exponential and Logarithm Functions Derivatives of Inverse Trig Functions Derivatives of Hyperbolic Functions Chain Rule Implicit Differentiation Related Rates Higher Order Derivatives Logarithmic Differentiation

The Definition of the Derivative Use the definition of the derivative to find the derivative of the following functions. 1. ( ) 6f x =

2. ( ) 3 14V t t= −

3. ( ) 2g x x=

4. ( ) 210 5Q t t t= + −

5. ( ) 24 9W z z z= −

6. ( ) 32 1f x x= −

7. ( ) 3 22 1g x x x x= − + −

8. ( ) 5R zz

=

9. ( ) 14

tV tt

+=

+

10. ( ) 3 4Z t t= −

11. ( ) 1 9f x x= −


Calculus I


Interpretations of the Derivative

For problems 1 and 2 use the graph of the function, ( )f x , estimate the value of ( )f a′ for the

given values of a. 1. (a) 2a = − (b) 3a =

2. (a) 1a = (b) 4a =

For problems 3 and 4 sketch the graph of a function that satisfies the given conditions. 3. ( )1 3f = , ( )1 1f ′ = , ( )4 5f = , ( )4 2f ′ = −

4. ( )3 5f − = , ( )3 2f ′ − = − , ( )1 2f = , ( )1 0f ′ = , ( )4 2f = − , ( )4 3f ′ = −


Calculus I


For problems 5 and 6 the graph of a function, ( )f x , is given. Use this to sketch the graph of the

derivative, ( )f x′ .

5.

6.

7. Answer the following questions about the function ( ) 24 9W z z z= − .

(a) Is the function increasing or decreasing at 1z = − ? (b) Is the function increasing or decreasing at 2z = ? (c) Does the function ever stop changing? If yes, at what value(s) of z does the function stop changing? 8. What is the equation of the tangent line to ( ) 3 14f x x= − at 8x = .


Calculus I


9. The position of an object at any time t is given by ( ) 14

ts tt

+=

+.

(a) Determine the velocity of the object at any time t. (b) Does the object ever stop moving? If yes, at what time(s) does the object stop moving?

10. What is the equation of the tangent line to ( ) 5f xx

= at 12

x = ?

11. Determine where, if anywhere, the function ( ) 3 22 1g x x x x= − + − stops changing.

12. Determine if the function ( ) 3 4Z t t= − increasing or decreasing at the given points.

(a) 5t = (b) 10t = (c) 300t = 13. Suppose that the volume of water in a tank for 0 6t≤ ≤ is given by ( ) 210 5Q t t t= + − .

(a) Is the volume of water increasing or decreasing at 0t = ? (b) Is the volume of water increasing or decreasing at 6t = ? (c) Does the volume of water ever stop changing? If yes, at what times(s) does the volume stop changing?

Differentiation Formulas For problems 1 – 12 find the derivative of the given function. 1. ( ) 36 9 4f x x x= − +

2. 4 22 10 13y t t t= − + 3. ( ) 7 74 3 9g z z z z−= − +

4. ( ) 4 3 29 8 12h y y y y− − −= − + +

5. 3 48 2y x x x= + −


Calculus I


6. ( ) 5 33 7 810 6 3f x x x x= − + −

7. ( ) 3 5

4 1 86

f tt t t

= − +

8. ( ) 4 103

6 1 18 3

R zz zz

= + −

9. ( )23 9z x x= −

10. ( ) ( ) ( )24 2g y y y y= − +

11. ( )34 7 8x xh x

x− +

=

12. ( )5 3

3

5 2y y yf yy

− +=

13. Determine where, if anywhere, the function ( ) 3 29 48 2f x x x x= + − + is not changing.

14. Determine where, if anywhere, the function 4 3 22 3y z z z= − − is not changing.

15. Find the tangent line to ( ) 16 4g x xx

= − at 4x = .

16. Find the tangent line to ( ) 4 67 8 2f x x x x−= + + at 1x = − .

17. The position of an object at any time t is given by ( ) 4 3 23 40 126 9s t t t t= − + − .

(a) Determine the velocity of the object at any time t. (b) Does the object ever stop changing? (c) When is the object moving to the right and when is the object moving to the left? 18. Determine where the function ( ) 3 4 56 40 5 4h z z z z= + − − is increasing and decreasing.

19. Determine where the function ( ) ( ) ( )21 2R x x x= + − is increasing and decreasing.


Calculus I


20. Determine where, if anywhere, the tangent line to ( ) 3 25f x x x x= − + is parallel to the line

4 23y x= + .

Product and Quotient Rule For problems 1 – 6 use the Product Rule or the Quotient Rule to find the derivative of the given function.

1. ( ) ( )( )2 3 24 8 12f t t t t t= − − +

2. ( )( )3 3 31 2y x x x−= + −

3. ( ) ( )( )2 2 31 2 3 5 8h z z z z z z= + + + −

4. ( )26

2xg x

x=

−

5. ( )4

2

32 1w wR ww

+=

+

6. ( ) 2

27 4

x xf xx x

+=

−

7. If ( )2 8f = − , ( )2 3f ′ = , ( )2 17g = and ( )2 4g′ = − determine the value of ( ) ( )2f g ′ .

8. If ( ) ( )3f x x g x= , ( )7 2g − = , ( )7 9g′ − = − determine the value of ( )7f ′ − .

9. Find the equation of the tangent line to ( ) ( ) ( )21 12 4f x x x= + − at 9x = .

10. Determine where ( )2

21 8x xf x

x−

=+

is increasing and decreasing.

11. Determine where ( ) ( )( )2 24 1 5V t t t= − + is increasing and decreasing.


Calculus I


Derivatives of Trig Functions For problems 1 – 3 evaluate the given limit.

1. ( )

0

sin 10limz

zz→

2. ( )( )0

sin 12lim

sin 5α

αα→

3. ( )

0

cos 4 1limx

xx→

−

For problems 4 – 10 differentiate the given function. 4. ( ) ( ) ( )2cos 6sec 3f x x x= − +

5. ( ) ( ) ( )10 tan 2cotg z z z= −

6. ( ) ( ) ( )tan secf w w w=

7. ( ) ( )3 2 sinh t t t t= −

8. ( )6 4 cscy x x= +

9. ( ) ( ) ( )1

2sin 4cosR t

t t=

−

10. ( ) ( )( )

tan1 cscv v

Z vv

+=

+

11. Find the tangent line to ( ) ( ) ( )tan 9cosf x x x= + at x π= .

12. The position of an object is given by ( ) ( )2 7coss t t= + determine all the points where the

object is not moving. 13. Where in the range [ ]2,7− is the function ( ) ( )4cosf x x x= − is increasing and

decreasing.


Calculus I


Derivatives of Exponential and Logarithm Functions For problems 1 – 6 differentiate the given function. 1. ( ) 2 8xxf x = −e

2. ( ) ( ) ( )34log lng t t t= −

3. ( ) ( )3 logwR w w=

4. ( )5 lnzy z z= −e

5. ( )1 y

yh y =− e

6. ( ) ( )1 5ln

tf tt

+=

7. Find the tangent line to ( ) 7 4x xf x = + e at 0x = .

8. Find the tangent line to ( ) ( ) ( )2ln logf x x x= at 2x = .

9. Determine if ( ) ttV t =e

is increasing or decreasing at the following points.

(a) 4t = − (b) 0t = (c) 10t = 10. Determine if ( ) ( ) ( )6 lnG z z z= − is increasing or decreasing at the following points.

(a) 1z = (b) 5z = (c) 20z =

Derivatives of Inverse Trig Functions For each of the following problems differentiate the given function. 1. ( ) ( ) ( )12cos 6cosT z z z−= +


Calculus I


2. ( ) ( ) ( )1 1csc 4cotg t t t− −= −

3. ( )6 15 secy x x−= −

4. ( ) ( ) ( )2 1sin tanf w w w w−= +

5. ( ) ( )1sin1

xh x

x

−

=+

Derivatives of Hyperbolic Functions For each of the following problems differentiate the given function. 1. ( ) ( ) ( ) ( )sinh 2cosh sechf x x x x= + −

2. ( ) ( ) ( )2tan cschR t t t t= +

3. ( ) ( )1

tanhzg z

z+

=

Chain Rule

For problems 1 – 26 differentiate the given function.

1. ( ) ( )426 7f x x x= +

2. ( ) ( ) 224 3 2g t t t−

= − +

3. 3 1 8y z= − 4. ( ) ( )csc 7R w w=

5. ( ) ( )( )2sin 3 tanG x x x= +


Calculus I


6. ( ) ( )tan 4 10h u u= +

7. ( ) 745 t tf t += + e

8. ( ) ( )1 cos xg x −= e

9. ( ) 1 62 zH z −=

10. ( ) ( )1tan 3 1u t t−= −

11. ( ) ( )2 3ln 1 5F y y y= − +

12. ( ) ( ) ( )( )ln sin cotV x x x= −

13. ( ) ( ) ( )6 6sin sinh z z z= +

14. ( ) 7 wS w w −= + e

15. ( ) ( )7 23 sin 6g z z z= − +

16. ( ) ( )( ) ( )104ln sin 3f x x x x= − −

17. ( ) 6 25h t t t t= −

18. ( ) ( )2 5lnq t t t=

19. ( ) ( ) ( )cos 3 sec 1g w w w= −

20. ( )

2

sin 31

ty

t=

+

21. ( ) ( )21

tan 12

xK x

x x

−+=

+e


Calculus I


22. ( ) ( )2cos xf x x= e

23. ( )5 tan 4z x x= +

24. ( ) ( )( )36 sin 2tf t t−= + −e

25. ( ) ( ) ( )( )102 1ln 1 tan 6g x x x−= + −

26. ( ) ( )4 2tan 1h z z= +

27. ( ) ( )( ) 123 12 sin 3f x x x−

= +

28. Find the tangent line to ( ) 24 2 6 xf x x −= − e at 2x = .

29. Determine where ( ) ( )34 2 8V z z z= − is increasing and decreasing.

30. The position of an object is given by ( ) ( )sin 3 2 4s t t t= − + . Determine where in the

interval [ ]0,3 the object is moving to the right and moving to the left.

31. Determine where ( ) 2 5 tA t t −= e is increasing and decreasing.

32. Determine where in the interval [ ]1, 20− the function ( ) ( )4 3ln 20 100f x x x= − − is

increasing and decreasing.

Implicit Differentiation For problems 1 – 3 do each of the following.

(a) Find y′ by solving the equation for y and differentiating directly. (b) Find y′ by implicit differentiation. (c) Check that the derivatives in (a) and (b) are the same.

1. 3 1xy

=


Calculus I


2. 2 3 4x y+ = 3. 2 2 2x y+ = For problems 4 – 9 find y′ by implicit differentiation. 4. 3 2 62 4y x y x+ − = 5. ( )2 47 sin 3 12y x y+ = −

6. ( )sinx y x− =e

7. 2 7 5 34 2 4x y x x y− = +

8. ( ) 22cos 2 1yx y x+ + =e

9. ( )2 4 2tan 3x y x y= +

For problems 10 & 11 find the equation of the tangent line at the given point.

10. 4 2 3x y+ = at ( )1, 2− .

11. 2 2 23xy y x= +e at ( )0,3 .

For problems 12 & 13 assume that ( )x x t= , ( )y y t= and ( )z z t= and differentiate the given

equation with respect to t. 12. 2 3 4 1x y z− + =

13. ( ) ( )2 3cos sin 4x y y z= +

Related Rates 1. In the following assume that x and y are both functions of t. Given 2x = − , 1y = and 4x′ = − determine y′ for the following equation.


Calculus I


2 2 3 4 46 2 yy x x −+ = − e 2. In the following assume that x, y and z are all functions of t. Given 4x = , 2y = − , 1z = ,

9x′ = and 3y′ = − determine z′ for the following equation.

( ) 3 2 2 21 5 3x y z y z x− + = + − 3. For a certain rectangle the length of one side is always three times the length of the other side.

(a) If the shorter side is decreasing at a rate of 2 inches/minute at what rate is the longer side decreasing?

(b) At what rate is the enclosed area decreasing when the shorter side is 6 inches long and is decreasing at a rate of 2 inches/minute?

4. A thin sheet of ice is in the form of a circle. If the ice is melting in such a way that the area of the sheet is decreasing at a rate of 0.5 m2/sec at what rate is the radius decreasing when the area of the sheet is 12 m2 ? 5. A person is standing 350 feet away from a model rocket that is fired straight up into the air at a rate of 15 ft/sec. At what rate is the distance between the person and the rocket increasing (a) 20 seconds after liftoff? (b) 1 minute after liftoff? 6. A plane is 750 meters in the air flying parallel to the ground at a speed of 100 m/s and is initially 2.5 kilometers away from a radar station. At what rate is the distance between the plane and the radar station changing (a) initially and (b) 30 seconds after it passes over the radar station? See the (probably bad) sketch below to help visualize the problem.

7. Two people are at an elevator. At the same time one person starts to walk away from the elevator at a rate of 2 ft/sec and the other person starts going up in the elevator at a rate of 7 ft/sec. What rate is the distance between the two people changing 15 seconds later? 8. Two people on bikes are at the same place. One of the bikers starts riding directly north at a rate of 8 m/sec. Five seconds after the first biker started riding north the second starts to ride directly east at a rate of 5 m/sec. At what rate is the distance between the two riders increasing 20 seconds after the second person started riding?


Calculus I


9. A light is mounted on a wall 5 meters above the ground. A 2 meter tall person is initially 10 meters from the wall and is moving towards the wall at a rate of 0.5 m/sec. After 4 seconds of moving is the tip of the shadow moving (a) towards or away from the person and (b) towards or away from the wall? 10. A tank of water in the shape of a cone is being filled with water at a rate of 12 m3/sec. The base radius of the tank is 26 meters and the height of the tank is 8 meters. At what rate is the depth of the water in the tank changing with the radius of the top of the water is 10 meters? 11. The angle of elevation is the angle formed by a horizontal line and a line joining the observer’s eye to an object above the horizontal line. A person is 500 feet way from the launch point of a hot air balloon. The hot air balloon is starting to come back down at a rate of 15 ft/sec. At what rate is the angle of elevation, θ , changing when the hot air balloon is 200 feet above the ground. See the (probably bad) sketch below to help visualize the angle of elevation if you are having trouble seeing it.

Higher Order Derivatives For problems 1 – 5 determine the fourth derivative of the given function. 1. ( ) 7 4 33 6 8 12 18h t t t t t= − + − +

2. ( ) 3 2 1V x x x x= − + −

3. ( ) 5 32

148

f x x xx

= − −

4. ( ) ( ) ( )37sin cos 1 2wf w w= + −

5. ( )5 48ln 2zy z−= +e

For problems 6 – 9 determine the second derivative of the given function.


Calculus I


6. ( ) ( )3sin 2 9g x x x= −

7. ( )3ln 7z x= −

8. ( )( )42

2

6 2Q v

v v=

+ −

9. ( ) ( )2cos 7H t t=

For problems 10 & 11 determine the second derivative of the given function. 10. 3 22 1 4x y y+ = − 11. 26 1y xy− =

Logarithmic Differentiation For problems 1 – 3 use logarithmic differentiation to find the first derivative of the given function.

1. ( ) ( )72 25 3 6 8 12f x x x x= − + −

2. ( )

( )

2

34

sin 3

6

z zy

z

+=

−

3. ( ) ( )3

24

5 8 1 9cos 4

10

t th t

t t

+ −=

+

For problems 4 & 5 find the first derivative of the given function.

4. ( ) ( )43 7 wg w w= −

5. ( ) ( ) ( )sin 282xxf x x= − e


Calculus I



CALCULUS I Solutions to Practice Problems

Derivatives

Paul Dawkins

Calculus I



The Definition of the Derivative .............................................................................................................. 2 Interpretations of the Derivative ........................................................................................................... 10 Differentiation Formulas ........................................................................................................................ 27 Product and Quotient Rule ..................................................................................................................... 40 Derivatives of Trig Functions ................................................................................................................ 46 Derivatives of Exponential and Logarithm Functions ......................................................................... 52 Derivatives of Inverse Trig Functions ................................................................................................... 56 Derivatives of Hyperbolic Functions ..................................................................................................... 58 Chain Rule ............................................................................................................................................... 59 Implicit Differentiation........................................................................................................................... 78 Related Rates .......................................................................................................................................... 88 Higher Order Derivatives ..................................................................................................................... 101 Logarithmic Differentiation ................................................................................................................. 108

Preface Here are the solutions to the practice problems for my Calculus I notes. Some solutions will have more or less detail than other solutions. The level of detail in each solution will depend up on several issues. If the section is a review section, this mostly applies to problems in the first chapter, there will probably not be as much detail to the solutions given that the problems really should be review. As the difficulty level of the problems increases less detail will go into the basics of the solution under the assumption that if you’ve reached the level of working the harder problems then you will probably already understand the basics fairly well and won’t need all the explanation. This document was written with presentation on the web in mind. On the web most solutions are broken down into steps and many of the steps have hints. Each hint on the web is given as a popup however in this document they are listed prior to each step. Also, on the web each step can be viewed individually by clicking on links while in this document they are all showing. Also, there are liable to be some formatting parts in this document intended for help in generating the web pages that haven’t been removed here. These issues may make the solutions a little difficult to follow at times, but they should still be readable.

Derivatives


Calculus I


The Definition of the Derivative 1. Use the definition of the derivative to find the derivative of, ( ) 6f x = Solution There really isn’t much to do for this problem other than to plug the function into the definition of the derivative and do a little algebra.

( ) ( ) ( )0 0 0 0

6 6 0lim lim lim lim0 0h h h h

f x h f xf x

h h h→ → → →

+ − −′ = = = = =

So, the derivative for this function is,

( ) 0f x′ = 2. Use the definition of the derivative to find the derivative of, ( ) 3 14V t t= − Step 1 First we need to plug the function into the definition of the derivative.

( ) ( ) ( ) ( ) ( )0 0

3 14 3 14lim limh h

V t h V t t h tV t

h h→ →

+ − − + − −′ = =

Make sure that you properly evaluate the first function evaluation. This is one of the more common errors that students make with these problems. Also watch for the parenthesis on the second function evaluation. You are subtracting off the whole function and so you need to make sure that you deal with the minus sign properly. Either put in the parenthesis as we’ve done here or make sure the minus sign get distributed through properly. This is another very common error and one that if you make will often make the problem impossible to complete. Step 2 Now all that we need to do is some quick algebra and we’ll be done.

( ) ( )0 0 0

3 14 14 3 14 14lim lim lim 14 14h h h

t h t hV th h→ → →

− − − + −′ = = = − = −

The derivative for this function is then,

( ) 14V t′ = −


Calculus I


3. Use the definition of the derivative to find the derivative of, ( ) 2g x x= Step 1 First we need to plug the function into the definition of the derivative.

( ) ( ) ( ) ( )2 2

0 0lim limh h

g x h g x x h xg x

h h→ →

+ − + −′ = =

Make sure that you properly evaluate the first function evaluation. This is one of the more common errors that students make with these problems. Step 2 Now all that we need to do is some quick algebra and we’ll be done.

( ) ( ) ( )2 2 2

0 0 0

22lim lim lim 2 2h h h

h x hx xh h xg x x h xh h→ → →

++ + −′ = = = + =


( ) 2g x x′ = 4. Use the definition of the derivative to find the derivative of, ( ) 210 5Q t t t= + − Step 1 First we need to plug the function into the definition of the derivative.

( ) ( ) ( ) ( ) ( ) ( )2 2

0 0

10 5 10 5lim limh h

t h t h t tQ t h Q tQ t

h h→ →

+ + − + − + −+ −′ = =

Make sure that you properly evaluate the first function evaluation. This is one of the more common errors that students make with these problems. Also watch for the parenthesis on the second function evaluation. You are subtracting off the whole function and so you need to make sure that you deal with the minus sign properly. Either put in the parenthesis as we’ve done here or make sure the minus sign get distributed through properly. This is another very common error and one that if you make will often make the problem impossible to complete.


Calculus I


Step 2 Now all that we need to do is some algebra (and it might get a little messy here, but that is somewhat common with these types of problems) and we’ll be done.

( )

( ) ( )

2 2 2

0

0 0

10 5 5 2 10 5lim

5 2lim lim 5 2 5 2

h

h h

t h t th h t tQ th

h t ht h t

h

→

→ →

+ + − − − − − +′ =

− −= = − − = −


( ) 5 2Q t t′ = − 5. Use the definition of the derivative to find the derivative of, ( ) 24 9W z z z= − Step 1 First we need to plug the function into the definition of the derivative.

( ) ( ) ( ) ( ) ( ) ( )2 2

0 0

4 9 4 9lim limh h

z h z h z zW z h W zW z

h h→ →

+ − + − −+ −′ = =

Make sure that you properly evaluate the first function evaluation. This is one of the more common errors that students make with these problems. Also watch for the parenthesis on the second function evaluation. You are subtracting off the whole function and so you need to make sure that you deal with the minus sign properly. Either put in the parenthesis as we’ve done here or make sure the minus sign get distributed through properly. This is another very common error and one that if you make will often make the problem impossible to complete. Step 2 Now all that we need to do is some algebra (and it might get a little messy here, but that is somewhat common with these types of problems) and we’ll be done.


Calculus I


( ) ( )

( ) ( )

2 2 2

0

2 2 2

0

0 0

4 2 9 9 4 9lim

4 8 4 9 9 4 9lim

8 4 9lim lim 8 4 9 8 9

h

h

h h

z zh h z h z zW z

hz zh h z h z z

hh z h

z h zh

→

→

→ →

+ + − − − +′ =

+ + − − − +=

+ −= = + − = −


( ) 8 9W z z′ = − 6. Use the definition of the derivative to find the derivative of, ( ) 32 1f x x= − Step 1 First we need to plug the function into the definition of the derivative.

( ) ( ) ( ) ( ) ( )3 3

0 0

2 1 2 1lim limh h

x h xf x h f xf x

h h→ →

+ − − −+ −′ = =

Make sure that you properly evaluate the first function evaluation. This is one of the more common errors that students make with these problems. Also watch for the parenthesis on the second function evaluation. You are subtracting off the whole function and so you need to make sure that you deal with the minus sign properly. Either put in the parenthesis as we’ve done here or make sure the minus sign get distributed through properly. This is another very common error and one that if you make will often make the problem impossible to complete. Step 2 Now all that we need to do is some algebra (and it might get a little messy here, but that is somewhat common with these types of problems) and we’ll be done.

( ) ( )

( ) ( )

3 2 2 3 3

0

3 2 2 3 3

0

2 22 2 2

0 0

2 3 3 1 2 1lim

2 6 6 2 1 2 1lim

6 6 2lim lim 6 6 2 6

h

h

h h

x x h xh h xf x

hx x h xh h x

hh x xh h

x xh h xh

→

→

→ →

+ + + − − +′ =

+ + + − − +=

+ += = + + =


Calculus I



( ) 26f x x′ = 7. Use the definition of the derivative to find the derivative of, ( ) 3 22 1g x x x x= − + − Step 1 First we need to plug the function into the definition of the derivative.

( ) ( ) ( ) ( ) ( ) ( )3 2 3 2

0 0

2 1 2 1lim limh h

x h x h x h x x xg x h g xg x

h h→ →

+ − + + + − − − + −+ −′ = =

Make sure that you properly evaluate the first function evaluation. This is one of the more common errors that students make with these problems. Also watch for the parenthesis on the second function evaluation. You are subtracting off the whole function and so you need to make sure that you deal with the minus sign properly. Either put in the parenthesis as we’ve done here or make sure the minus sign get distributed through properly. This is another very common error and one that if you make will often make the problem impossible to complete. Step 2 Now all that we need to do is some algebra (and it will get a little messy here, but that is somewhat common with these types of problems) and we’ll be done.

( ) ( ) ( )

( )

( )

3 2 2 3 2 2 3 2

0

3 2 2 3 2 2 3 2

0

2 2

0

2 2 2

0

3 3 2 2 1 2 1lim

3 3 2 4 2 1 2 1lim

3 3 4 2 1lim

lim 3 3 4 2 1 3 4 1

h

h

h

h

x x h xh h x xh h x h x x xg x

hx x h xh h x xh h x h x x x

hh x xh h x h

hx xh h x h x x

→

→

→

→

+ + + − + + + + − − − + −′ =

+ + + − − − + + − − + − +=

+ + − − +=

= + + − − + = − +


( ) 23 4 1g x x x′ = − +


Calculus I


8. Use the definition of the derivative to find the derivative of,

( ) 5R zz

=

Step 1 First we need to plug the function into the definition of the derivative.

( ) ( ) ( )0 0

1 5 5lim limh h

R z h R zR z

h h z h z→ →

+ − ′ = = − +

Make sure that you properly evaluate the first function evaluation. This is one of the more common errors that students make with these problems. Also note that in order to make the problem a little easier to read rewrote the rational expression in the definition a little bit. This doesn’t need to be done, but will make things a little nicer to look at. Step 2 Next we need to combine the two rational expressions into a single rational expression.

( ) ( )( )0

5 51limh

z z hR z

h z z h→

− +′ = +

Step 3 Now all that we need to do is some algebra and we’ll be done.

( ) ( ) ( ) ( ) 20 0 0

1 5 5 5 1 5 5 5lim lim limh h h

z z h hR zh z z h h z z h z z h z→ → →

− − − −′ = = = = − + + +


( ) 2

5R zz

′ = −


( ) 14

tV tt

+=

+

Step 1 First we need to plug the function into the definition of the derivative.

( ) ( ) ( )0 0

1 1 1lim lim4 4h h

V t h V t t h tV th h t h t→ →

+ − + + + ′ = = − + + +


Calculus I


Make sure that you properly evaluate the first function evaluation. This is one of the more common errors that students make with these problems. Also note that in order to make the problem a little easier to read rewrote the rational expression in the definition a little bit. This doesn’t need to be done, but will make things a little nicer to look at. Step 2 Next we need to combine the two rational expressions into a single rational expression.

( ) ( )( ) ( )( )( ) ( )0

1 4 1 41lim4 4h

t h t t t hV t

h t h t→

+ + + − + + +′ = + + +

Step 3 Now all that we need to do is some algebra (and it will get a little messy here, but that is somewhat common with these types of problems) and we’ll be done.

( ) ( )( )( )

( )( )

( )( ) ( ) ( ) ( )

2 2

0

2 2

0

20 0

5 4 4 5 41lim4 4

1 5 4 4 5 4lim4 4

1 3 3 3lim lim4 4 4 4 4

h

h

h h

t th t h t th t hV t

h t h t

t th t h t th t hh t h t

hh t h t t h t t

→

→

→ →

+ + + + − + + + +′ =

+ + + + + + + − − − − −

= + + +

= = = + + + + + + +


( )( )2

34

V tt

′ =+


( ) 3 4Z t t= − Step 1 First we need to plug the function into the definition of the derivative.

( ) ( ) ( ) ( )0 0

3 4 3 4lim limh h

t h tZ t h Z tZ t

h h→ →

+ − − −+ −′ = =


Calculus I


Make sure that you properly evaluate the first function evaluation. This is one of the more common errors that students make with these problems. Step 2 Next we need to rationalize the numerator.

( )( )( ) ( )( )

( )( )0

3 4 3 4 3 4 3 4lim

3 4 3 4h

t h t t h tZ t

h t h t→

+ − − − + − + −′ =

+ − + −


( ) ( ) ( )( )( ) ( )( )

( )( ) ( )

0 0

0 0

3 4 3 4 3 3 4 3 4lim lim3 4 3 4 3 4 3 4

3 3 3lim lim2 3 43 4 3 43 4 3 4

h h

h h

t h t t h tZ th t h t h t h t

htt h th t h t

→ →

→ →

+ − − − + − − +′ = =+ − + − + − + −

= = =−+ − + −+ − + −

Be careful when multiplying out the numerator here. It is easy to lose track of the minus sign (or parenthesis for that matter) on the second term. This is a very common mistake that students make. The derivative for this function is then,

( ) 32 3 4

Z tt

′ =−


( ) 1 9f x x= − Step 1 First we need to plug the function into the definition of the derivative.

( ) ( ) ( ) ( )0 0

1 9 1 9lim limh h

x h xf x h f xf x

h h→ →

− + − −+ −′ = =

Make sure that you properly evaluate the first function evaluation. This is one of the more common errors that students make with these problems. Step 2


Calculus I


Next we need to rationalize the numerator.

( )( )( ) ( )( )

( )( )0

1 9 1 9 1 9 1 9lim

1 9 1 9h

x h x x h xf x

h x h x→

− + − − − + + −′ =

− + + −


( ) ( ) ( )( )( ) ( )( )

( )( ) ( )

0 0

0 0

1 9 1 9 1 9 9 1 9lim lim1 9 1 9 1 9 1 9

9 9 9lim lim2 1 91 9 1 91 9 1 9

h h

h h

x h x x h xf xh x h x h x h x

hxx h xh x h x

→ →

→ →

− + − − − − − +′ = =− + + − − + + −

− − −= = =

−− + + −− + + −

Be careful when multiplying out the numerator here. It is easy to lose track of the minus sign (or parenthesis for that matter) on the second term. This is a very common mistake that students make. The derivative for this function is then,

( ) 92 1 9

f xx

−′ =−

Interpretations of the Derivative 1. Use the graph of the function, ( )f x , estimate the value of ( )f a′ for

(a) 2a = − (b) 3a =


Calculus I


Solution Hint : Remember that one of the interpretations of the derivative is the slope of the tangent line to the function. (a) 2a = − Step 1 Given that one of the interpretations of the derivative is that it is the slope of the tangent line to the function at a particular point let’s first sketch in a tangent line at the point on the graph.

Step 2 The function is clearly decreasing here and so we know that the derivative at this point will be negative. Now, from this sketch of the tangent line it looks like if we run over 1 we go down 4 and so we can estimate that,

( )2 4f ′ − = −

(b) 3a =


Calculus I


Step 1 Given that one of the interpretations of the derivative is that it is the slope of the tangent line to the function at a particular point. Let’s first sketch in a tangent line at the point.

Step 2 The function is clearly increasing here and so we know that the derivative at this point will be positive. Now, from this sketch of the tangent line it looks like if we run over 1 we go up 2 and so we can estimate that,

( )3 2f ′ =

2. Use the graph of the function, ( )f x , estimate the value of ( )f a′ for

(a) 1a = (b) 4a =

Solution


Calculus I


Hint : Remember that one of the interpretations of the derivative is the slope of the tangent line to the function. (a) 1a = Step 1 Given that one of the interpretations of the derivative is that it is the slope of the tangent line to the function at a particular point let’s first sketch in a tangent line at the point on the graph.

Step 2 The function is clearly decreasing here and so we know that the derivative at this point will be positive. Now, from this sketch of the tangent line it looks like if we run over 1 we go up 1 and so we can estimate that,

( )1 1f ′ =

(b) 4a = Step 1 Given that one of the interpretations of the derivative is that it is the slope of the tangent line to the function at a particular point. Let’s first sketch in a tangent line at the point.


Calculus I


Step 2 The function is clearly decreasing here and so we know that the derivative at this point will be positive. Now, from this sketch of the tangent line it looks like if we run over 1 we go up 5 and so we can estimate that,

( )4 5f ′ =

3. Sketch the graph of a function that satisfies ( )1 3f = , ( )1 1f ′ = , ( )4 5f = , ( )4 2f ′ = − . Solution Hint : Remember that one of the interpretations of the derivative is the slope of the tangent line to the function. Step 1 First, recall that one of the interpretations of the derivative is that it is the slope of the tangent line to the function at a particular point. So, let’s start off with a graph that has the given points on it and a sketch of a tangent line at the points whose slope is the value of the derivative at the points.


Calculus I


Step 2 Now, all that we need to do is sketch in a graph that goes through the indicated points and at the same time it must be parallel to the tangents that we sketched. There are many possible sketches that we can make here and so don’t worry if your sketch is not the same as the one here. This is just one possible sketch that meets the given conditions.

While, it’s not really needed here is a sketch of the function without all the extra bits that we put in to help with the sketch.


Calculus I


4. Sketch the graph of a function that satisfies ( )3 5f − = , ( )3 2f ′ − = − , ( )1 2f = , ( )1 0f ′ = ,

( )4 2f = − , ( )4 3f ′ = − . Solution Hint : Remember that one of the interpretations of the derivative is the slope of the tangent line to the function. Step 1 First, recall that one of the interpretations of the derivative is that it is the slope of the tangent line to the function at a particular point. So, let’s start off with a graph that has the given points on it and a sketch of a tangent line at the points whose slope is the value of the derivative at the points.

Step 2


Calculus I


Now, all that we need to do is sketch in a graph that goes through the indicated points and at the same time it must be parallel to the tangents that we sketched. There are many possible sketches that we can make here and so don’t worry if your sketch is not the same as the one here. This is just one possible sketch that meets the given conditions.

While, it’s not really needed here is a sketch of the function without all the extra bits that we put in to help with the sketch.

5. Below is the graph of some function, ( )f x . Use this to sketch the graph of the derivative,

( )f x′ .


Calculus I


Solution Hint 1 : Where will the derivative be zero? This gives us a couple of starting points for our sketch. Step 1 From the graph of the function it is pretty clear that we will have horizontal tangent lines at

2x = − , 1x = and 5x = . Because we will have horizontal tangents here we also know that the derivative at these points must be zero. Therefore, we know the following derivative evaluations. ( ) ( ) ( )2 0 1 0 5 0f f f′ ′ ′− = = = Hint 2 : Recall that the derivative can also be used to tell us where the function is increasing and decreasing. Knowing this we can use the graph to determine where the derivative will be positive and where it will be negative. Step 2 The points we found above break the x-axis up into regions where the function is increasing and decreasing. Recall that if the derivative is positive then the function is increasing and likewise if the derivative is negative then the function is decreasing. Using these ideas we can easily identify the sign of the derivative on each of the regions. Doing this gives,

( )( )( )( )

2 0

2 1 0

1 5 0

5 0

x f x

x f x

x f x

x f x

′< − >

′− < < <

′< < >

′> <

Hint 3 : At this point all we have to do is try and put all this together and come up with a sketch of the derivative.


Calculus I


Step 3 This is the tricky part of this problem. In the range 2x < − we know the derivative must be positive and that it must be zero at 2x = − so it makes sense that just to the left of 2x = − the derivative must be decreasing. In the range 2 1x− < < we know that the derivative will be negative and that it will be zero at the endpoints of the range. So, to the right of 2x = − the derivative will have to be decreasing (goes from zero to a negative number). Likewise, to the left of 1x = the derivative will have to be increasing (goes from a negative number to zero). Note that we don’t really know just how the derivative will behave everywhere in this range, but we can use the general behavior near the endpoints and go with the simplest way to connect the two up to get an idea of what the derivative should look like. Following similar reasoning we can see that the derivative should be increasing just to the right of

1x = (goes from zero to a positive number), decreasing just to the left of 5x = (goes from a positive number to zero) and decreasing just to the right of 5x = (goes from zero to a negative number). Step 4 So, putting all of this together here is a sketch of the derivative. Note that we included the a scale on the vertical axis if you would like to try and estimate some specific values of the derivative as we did in Example 4 of this section.


Calculus I


6. Below is the graph of some function, ( )f x . Use this to sketch the graph of the derivative,

( )f x′ .

Solution Hint 1 : Because the derivative of a function is also the slope of the tangent line. We can therefore determine actual values of the derivative at almost every spot. Step 1 Because the three portions of the function are actually lines and the tangent line to a line would just be the line itself we can easily compute the derivative on each portion of the curve. On each of the portions we can use the grid included on the graph to compute the slope of each part. Knowing the slope of the graph on each portion will in turn tell us the slope of the tangent line for each portion. This in turn tells us that the derivative on each of the three portions is then,

( )

( )

( )

1 121 23

2 2

x f x

x f x

x f x

′< − = −

′− < < =

′> = −

Hint 2 : What is the derivative at the “sharp points”? Step 2 Recall Example 4 from the previous section. In that example we showed that the derivative of the absolute value function does not exist at 0x = . The limit on the left side of 0x = (which gives the slope of the line on the left) and the limit on the right side of 0x = (which gives the slope of the line on the right) were different and so the overall limit did not exist. This in turn tells us that the derivative doesn’t exist at that point.


Calculus I


Here we have the same problem. We’ll leave it to you to verify that the right and left handed limits at 1x = − are not the same and so the derivative does not exist at 1x = − . Likewise, the derivative does not exist at 2x = . There will therefore be open dots on the graph at these two points. Step 3 Here is the sketch of the derivative of this function.

7. Answer the following questions about the function ( ) 24 9W z z z= − .

(a) Is the function increasing or decreasing at 1z = − ? (b) Is the function increasing or decreasing at 2z = ? (c) Does the function ever stop changing? If yes, at what value(s) of z does the function stop changing? (a) Is the function increasing or decreasing at 1z = − ? We know that the derivative of a function gives us the rate of change of the function and so we’ll first need the derivative of this function. We computed this derivative in Problem 5 from the previous section and so we won’t show the work here. If you need the practice you should go back and redo that problem before proceeding. So, from our previous work we know that the derivative is, ( ) 8 9W z z′ = − Now all that we need to do is to compute : ( )1 17W ′ − = − . This is negative and so we know that

the function must be decreasing at 1z = − .


Calculus I


(b) Is the function increasing or decreasing at 2z = ? Again, all we need to do is compute a derivative and since we’ve got the derivative written down in the first part there’s no reason to redo that here. The evaluation is : ( )2 7W ′ = . This is positive and so we know that the function must be

increasing at 2z = . (c) Does the function ever stop changing? If yes, at what value(s) of z does the function stop changing? Here all that we’re really asking is if the derivative is ever zero. So we need to solve,

( ) 90 8 9 08

W z z z′ = → − = ⇒ =

So, the function will stop changing at 98

z = .

8. What is the equation of the tangent line to ( ) 3 14f x x= − at 8x = .

Solution We know that the derivative of a function gives us the slope of the tangent line and so we’ll first need the derivative of this function. We computed this derivative in Problem 2 from the previous section and so we won’t show the work here. If you need the practice you should go back and redo that problem before proceeding. Note that we did use a different set of letters in the previous problem, but the work is identical. So, from our previous work (with a corresponding change of variables) we know that the derivative is, ( ) 14f x′ = − This tells us that the slope of the tangent line at 8x = is then : ( )8 14m f ′= = − . We also know

that a point on the tangent line is : ( )( ) ( )8, 8 8, 109f = − .

The tangent line is then, ( )109 14 8 3 14y x x= − − − = −


Calculus I


Note that, in this case the tangent is the same as the function. This should not be surprising however as the function is a line and so any tangent line (i.e. parallel line) will in fact be the same as the line itself.

9. The position of an object at any time t is given by ( ) 14

ts tt

+=

+.

(a) Determine the velocity of the object at any time t. (b) Does the object ever stop moving? If yes, at what time(s) does the object stop moving? (a) Determine the velocity of the object at any time t. We know that the derivative of a function gives is the velocity of the object and so we’ll first need the derivative of this function. We computed this derivative in Problem 9 from the previous section and so we won’t show the work here. If you need the practice you should go back and redo that problem before proceeding. Note that we did use a different letter for the function in the previous problem, but the work is identical. So, from our previous work we know that the derivative is,

( )( )2

34

s tt

′ =+

(b) Does the object ever stop moving? If yes, at what time(s) does the object stop moving? We know that the object will stop moving if the velocity (i.e. the derivative) is zero. In this case the derivative is a rational expression and clearly the numerator will never be zero. Therefore, the derivative will not be zero and therefore the object never stops moving.

10. What is the equation of the tangent line to ( ) 5f xx

= at 12

x = ?

Solution We know that the derivative of a function gives us the slope of the tangent line and so we’ll first need the derivative of this function. We computed this derivative in Problem 8 from the previous section and so we won’t show the work here. If you need the practice you should go back and redo that problem before proceeding.


Calculus I


Note that we did use a different set of letters in the previous problem, but the work is identical. So, from our previous work (with a corresponding change of variables) we know that the derivative is,

( ) 2

5f xx

′ = −

This tells us that the slope of the tangent line at 12

x = is then : 1 202

m f ′= = −

. We also

know that a point on the tangent line is : 1 1 1, ,102 2 2

f = .

The tangent line is then,

110 20 20 202

y x x = − − = −

11. Determine where, if anywhere, the function ( ) 3 22 1g x x x x= − + − stops changing.

Solution We know that the derivative of a function gives us the rate of change of the function and so we’ll first need the derivative of this function. We computed this derivative in Problem 7 from the previous section and so we won’t show the work here. If you need the practice you should go back and redo that problem before proceeding. From our previous work (with a corresponding change of variables) we know that the derivative is, ( ) 23 4 1g x x x′ = − + If the function stops changing at a point then the derivative will be zero at that point. So, to determine if we function stops changing we will need to solve,

( )

( ) ( )

2

0

3 4 1 013 1 1 0 , 13

g x

x x

x x x x

′ =

− + =

− − = ⇒ = =

So, the function will stop changing at 13

x = and 1x = .


Calculus I


12. Determine if the function ( ) 3 4Z t t= − increasing or decreasing at the given points.

(a) 5t = (b) 10t = (c) 300t = (a) 5t = We know that the derivative of a function gives us the rate of change of the function and so we’ll first need the derivative of this function. We computed this derivative in Problem 10 from the previous section and so we won’t show the work here. If you need the practice you should go back and redo that problem before proceeding. So, from our previous work we know that the derivative is,

( ) 32 3 4

Z tt

′ =−

Now all that we need to do is to compute : ( ) 352 11

Z ′ = . This is positive and so we know that

the function must be increasing at 5t = . (b) 10t = Again, all we need to do is compute a derivative and since we’ve got the derivative written down in the first part there’s no reason to redo that here.

The evaluation is : ( ) 3102 26

Z ′ = . This is positive and so we know that the function must be

increasing at 10t = . (c) 300t = Again, all we need to do is compute a derivative and since we’ve got the derivative written down in the first part there’s no reason to redo that here.

The evaluation is : ( ) 33002 896

Z ′ = . This is positive and so we know that the function must

be increasing at 300t = . Final Note


Calculus I


As a final note to all the parts of this problem let’s notice that we did not really need to do any evaluations. Because we know that square roots will always be positive it is clear that the derivative will always be positive regardless of the value of t we plug in. 13. Suppose that the volume of water in a tank for 0 6t≤ ≤ is given by ( ) 210 5Q t t t= + − .

(a) Is the volume of water increasing or decreasing at 0t = ? (b) Is the volume of water increasing or decreasing at 6t = ? (c) Does the volume of water ever stop changing? If yes, at what times(s) does the volume stop changing? (a) Is the volume of water increasing or decreasing at 0t = ? We know that the derivative of a function gives us the rate of change of the function and so we’ll first need the derivative of this function. We computed this derivative in Problem 4 from the previous section and so we won’t show the work here. If you need the practice you should go back and redo that problem before proceeding. So, from our previous work we know that the derivative is, ( ) 5 2Q t t′ = − Now all that we need to do is to compute : ( )0 5Q′ = . This is positive and so we know that the

volume of water in the tank must be increasing at 0t = . (b) Is the volume of water increasing or decreasing at 6t = ? Again, all we need to do is compute a derivative and since we’ve got the derivative written down in the first part there’s no reason to redo that here. The evaluation is : ( )6 7Q′ = − . This is negative and so we know that the volume of water in the

tank must be decreasing at 6t = . (c) Does the volume of water ever stop changing? If yes, at what times(s) does the volume stop changing? Here all that we’re really asking is if the derivative is ever zero. So we need to solve,

( ) 50 5 2 02

Q t t t′ = → − = ⇒ =


Calculus I


So, the volume of water will stop changing at 52

.

Differentiation Formulas 1. Find the derivative of ( ) 36 9 4f x x x= − + .

Solution There isn’t much to do here other than take the derivative using the rules we discussed in this section.

( ) 218 9f x x′ = − 2. Find the derivative of 4 22 10 13y t t t= − + . Solution There isn’t much to do here other than take the derivative using the rules we discussed in this section.

38 20 13dy t tdt

= − +

3. Find the derivative of ( ) 7 74 3 9g z z z z−= − + .


( ) 6 828 21 9g z z z−′ = + +


Calculus I


4. Find the derivative of ( ) 4 3 29 8 12h y y y y− − −= − + + .


( ) 5 4 34 27 16h y y y y− − −′ = − + −

5. Find the derivative of 3 48 2y x x x= + − . Solution There isn’t much to do here other than take the derivative using the rules we discussed in this section. Remember that you’ll need to convert the roots to fractional exponents before you start taking the derivative. Here is the rewritten function.

11 132 48 2y x x x= + −

The derivative is,

21 332 41 8 1

2 3 2dy x x xdx

−− −= + −

6. Find the derivative of ( ) 5 33 7 810 6 3f x x x x= − + − .

Solution There isn’t much to do here other than take the derivative using the rules we discussed in this section. Remember that you’ll need to convert the roots to fractional exponents before you start taking the derivative. Here is the rewritten function.

( ) ( ) ( ) ( )3 871 1 1

3 7 8 5 325 2 310 6 3 10 6 3f x x x x x x x= − + − = − + −


Calculus I


The derivative is,

( )2 5 2 55 55 3 5 32 23 7 8 710 6 6 16

5 2 3 2f x x x x x x x

− − ′ = − + = − +

7. Find the derivative of ( ) 3 5

4 1 86

f tt t t

= − + .

Solution There isn’t much to do here other than take the derivative using the rules we discussed in this section. Remember that you’ll need to rewrite the terms so that each of the t’s are in the numerator with negative exponents before taking the derivative. Here is the rewritten function.

( ) 1 3 514 86

f t t t t− − −= − +

The derivative is,

( ) 2 4 614 402

f t t t t− − −′ = − + −

8. Find the derivative of ( ) 4 103

6 1 18 3

R zz zz

= + − .

Solution There isn’t much to do here other than take the derivative using the rules we discussed in this section. Remember that you’ll need to rewrite the terms so that each of the z’s are in the numerator with negative exponents and rewrite the root as a fractional exponent before taking the derivative. Here is the rewritten function.

( )3

4 102 1 168 3

R z z z z− − −= + −

The derivative is,


Calculus I


( ) ( ) ( )5 5

5 11 5 112 23 1 1 1 106 4 10 92 8 3 2 3

R z z z z z z z− −− − − − ′ = − + − − − = − − +

9. Find the derivative of ( )23 9z x x= − .

Solution There isn’t much to do here other than take the derivative using the rules we discussed in this section. Remember that in order to do this derivative we’ll first need to multiply the function out before we take the derivative. Here is the rewritten function. 33 9z x x= − The derivative is,

29 9dz xdx

= −

10. Find the derivative of ( ) ( )( )24 2g y y y y= − + .

Solution There isn’t much to do here other than take the derivative using the rules we discussed in this section. Remember that in order to do this derivative we’ll first need to multiply the function out before we take the derivative. Here is the rewritten function. ( ) 3 22 8g y y y y= − − The derivative is,

( ) 23 4 8g y y y′ = − −


Calculus I


11. Find the derivative of ( )34 7 8x xh x

x− +

= .

Solution There isn’t much to do here other than take the derivative using the rules we discussed in this section. Remember that in order to do this derivative we’ll first need to divide the function out and simplify before we take the derivative. Here is the rewritten function.

( )3

2 14 7 8 4 7 8x xh x x xx x x

−= − + = − +

The derivative is,

( ) 28 8h x x x−′ = −

12. Find the derivative of ( )5 3

3

5 2y y yf yy

− += .

Solution There isn’t much to do here other than take the derivative using the rules we discussed in this section. Remember that in order to do this derivative we’ll first need to divide the function out and simplify before we take the derivative. Here is the rewritten function.

( )5 3

2 23 3 3

5 2 5 2y y yf y y yy y y

−= − + = − +

The derivative is,

( ) 32 4f y y y−′ = − 13. Determine where, if anywhere, the function ( ) 3 29 48 2f x x x x= + − + is not changing.

Hint : Recall the various interpretations of the derivative. One of them is exactly what we need to do this problem.


Calculus I


Solution Step 1 Recall that one of the interpretations of the derivative is that it gives the rate of change of the function. So, the function won’t be changing if its rate of change is zero and so all we need to do is find the derivative and set it equal to zero to determine where the rate of change is zero and hence the function will not be changing. First the derivative, and we’ll do a little factoring while we are at it.

( ) ( ) ( ) ( )2 23 18 48 3 6 16 3 8 2f x x x x x x x′ = + − = + − = + − Step 2 Now all that we need to do is set this equation to zero and solve. ( ) ( ) ( )0 3 8 2 0f x x x′ = ⇒ + − = We can easily see from this that the derivative will be zero at 8x = − and 2x = . The function therefore not be changing at,

8 and 2x x= − = 14. Determine where, if anywhere, the function 4 3 22 3y z z z= − − is not changing. Hint : Recall the various interpretations of the derivative. One of them is exactly what we need to do this problem. Solution Step 1 Recall that one of the interpretations of the derivative is that it gives the rate of change of the function. So, the function won’t be changing if its rate of change is zero and so all we need to do is find the derivative and set it equal to zero to determine where the rate of change is zero and hence the function will not be changing. First the derivative, and we’ll do a little factoring while we are at it.

( )3 2 28 3 6 8 3 6dy z z z z z zdz

= − − = − −

Step 2


Calculus I


Now all that we need to do is set this equation to zero and solve.

( )2 2

0

8 3 6 0 0, 8 3 6 0

dydz

z z z z z z

=

− − = → = − − =

We can easily see from this that the derivative will be zero at 0z = , however, because the quadratic doesn’t factor we’ll need to use the quadratic formula to determine where, if anywhere, that will be zero.

( ) ( ) ( )

( )

23 3 4 8 6 3 2012 8 16

z± − − − ±

= =

The function therefore not be changing at,

3 201 3 2010 1.07359 0.6985916 16

z z z+ −= = = = = −

15. Find the tangent line to ( ) 16 4g x xx

= − at 4x = .

Hint : Recall the various interpretations of the derivative. One of them will help us do this problem. Solution Step 1 Recall that one of the interpretations of the derivative is that it gives slope of the tangent line to the graph of the function. So, we’ll need the derivative of the function. However before doing that we’ll need to do a little rewrite. Here is that work as well as the derivative.

( ) ( )1 1

1 22 22

16 216 4 16 2g x x x g x x xx x

−− −′= − ⇒ = − − = − −

Note that we rewrote the derivative back into rational expressions with roots to help with the evaluation. Step 2 Next we need to evaluate the function and derivative at 4x = .


Calculus I


( ) ( ) 2

16 16 24 4 4 4 4 24 4 4

g g′= − = − = − − −

Step 3 Now all that we need to do is write down the equation of the tangent line.

( ) ( ) ( ) ( )4 4 4 4 2 4 2 4y g g x x y x′= + − = − − − → = − + 16. Find the tangent line to ( ) 4 67 8 2f x x x x−= + + at 1x = − .

Hint : Recall the various interpretations of the derivative. One of them will help us do this problem. Solution Step 1 Recall that one of the interpretations of the derivative is that it gives slope of the tangent line to the graph of the function. So, we’ll need the derivative of the function.

( ) 3 7 37

4828 48 2 28 2f x x x xx

−′ = − + = − +

Note that we rewrote the derivative back into rational expressions help a little with the evaluation. Step 2 Next we need to evaluate the function and derivative at 1x = − .

( ) ( )1 7 8 2 13 1 28 48 2 22f f ′− = + − = − = − + + = Step 3 Now all that we need to do is write down the equation of the tangent line.

( ) ( )( ) ( )1 1 1 13 22 1 22 35y f f x x y x′= − + − + = + + → = + 17. The position of an object at any time t is given by ( ) 4 3 23 40 126 9s t t t t= − + − .

(a) Determine the velocity of the object at any time t. (b) Does the object ever stop changing? (c) When is the object moving to the right and when is the object moving to the left?


Calculus I


Solution Hint : Recall the various interpretations of the derivative. One of them is exactly what we need for this part. (a) Determine the velocity of the object at any time t. Recall that one of the interpretations of the derivative is that it gives the velocity of an object if we know the position function of the object. We’ve been given the position function of the object and so all we need to do is find its derivative and we’ll have the velocity of the object at any time t. The velocity of the object is then,

( ) ( )( )3 212 120 252 12 3 7s t t t t t t t′ = − + = − − Note that the derivative was factored for later parts and doesn’t really need to be done in general. Hint : If the object isn’t moving what is the velocity? (b) Does the object ever stop changing? The object will not be moving if the velocity is ever zero and so all we need to do is set the derivative equal to zero and solve. ( ) ( ) ( )0 12 3 7 0s t t t t′ = ⇒ − − = From this it is pretty easy to see that the derivative will be zero, and hence the object will not be moving, at,

0 3 7t t t= = = Hint : How does the direction (right vs. left) of movement relate to the sign (positive or negative) of the derivative? (c) When is the object moving to the right and when is the object moving to the left? To answer this part all we need to know is where the derivative is positive (and hence the object is moving to the right) or negative (and hence the object is moving to the left). Because the derivative is continuous we know that the only place it can change sign is where the derivative is zero. So, as we did in this section a quick number line will give us the sign of the derivative for the various intervals.


Calculus I


Here is the number line for this problem.

From this we get the following right/left movement information.

Moving Right : 0 3, 7Moving Left : 0, 3 7

t tt t

< < < < ∞− ∞ < < < <

Note that depending upon your interpretation of t as time you may or may not have included the interval 0t−∞ < < in the “Moving Left” portion. 18. Determine where the function ( ) 3 4 56 40 5 4h z z z z= + − − is increasing and decreasing.

Solution Hint : Recall the various interpretations of the derivative. One of them is exactly what we need to get the problem started. Step 1 Recall that one of the interpretations of the derivative is that it gives the rate of change of the function. Since we are talking about where the function is increasing and decreasing we are clearly talking about the rate of change of the function. So, we’ll need the derivative.

( ) ( ) ( )2 3 4 2120 20 20 20 3 2h z z z z z z z′ = − − = − + − Note that the derivative was factored for later steps and doesn’t really need to be done in general. Hint : Where is the function not changing? Step 2


Calculus I


Next, we need to know where the function is not changing and so all we need to do is set the derivative equal to zero and solve. ( ) ( ) ( )20 20 3 2 0h z z z z′ = ⇒ − + − = From this it is pretty easy to see that the derivative will be zero, and hence the function will not be moving, at,

0 3 2z z z= = − = Hint : How does the increasing/decreasing behavior of the function relate to the sign (positive or negative) of the derivative? Step 3 To get the answer to this problem all we need to know is where the derivative is positive (and hence the function is increasing) or negative (and hence the function is decreasing). Because the derivative is continuous we know that the only place it can change sign is where the derivative is zero. So, as we did in this section a quick number line will give us the sign of the derivative for the various intervals. Here is the number line for this problem.

From this we get the following increasing/decreasing information.

Increasing : 3 0, 0 2Decreasing : 3, 2

z zz z

− < < < <− ∞ < < − < < ∞

19. Determine where the function ( ) ( ) ( )21 2R x x x= + − is increasing and decreasing.

Solution Hint : Recall the various interpretations of the derivative. One of them is exactly what we need to get the problem started.


Calculus I


Step 1 Recall that one of the interpretations of the derivative is that it gives the rate of change of the function. Since we are talking about where the function is increasing and decreasing we are clearly talking about the rate of change of the function. So, we’ll need the derivative. First however we’ll need to multiply out the function so we can actually take the derivative. Here is the rewritten function and the derivative.

( ) ( ) ( )3 2 23 4 3 6 3 2R x x x R x x x x x′= − + = − = − Note that the derivative was factored for later steps and doesn’t really need to be done in general. Hint : Where is the function not changing? Step 2 Next, we need to know where the function is not changing and so all we need to do is set the derivative equal to zero and solve. ( ) ( )0 3 2 0R x x x′ = ⇒ − = From this it is pretty easy to see that the derivative will be zero, and hence the function will not be moving, at,

0 2x x= = Hint : How does the increasing/decreasing behavior of the function relate to the sign (positive or negative) of the derivative? Step 3 To get the answer to this problem all we need to know is where the derivative is positive (and hence the function is increasing) or negative (and hence the function is decreasing). Because the derivative is continuous we know that the only place it can change sign is where the derivative is zero. So, as we did in this section a quick number line will give us the sign of the derivative for the various intervals. Here is the number line for this problem.


Calculus I



Increasing : 0, 2Decreasing : 0 2

x zz

− ∞ < < < < ∞< <

20. Determine where, if anywhere, the tangent line to ( ) 3 25f x x x x= − + is parallel to the line

4 23y x= + . Solution Step 1 The first thing that we’ll need of course is the slope of the tangent line. So, all we need to do is take the derivative of the function.

( ) 23 10 1f x x x′ = − + Hint : What is the relationship between the slope of two parallel lines? Step 2 Two lines that are parallel will have the same slope and so all we need to do is determine where the slope of the tangent line will be 4, the slope of the given line. In other words, we’ll need to solve, ( ) 2 24 3 10 1 4 3 10 3 0f x x x x x′ = → − + = → − − = This quadratic doesn’t factor and so a quick use of the quadratic formula will solve this for us.

10 136 10 2 34 5 346 6 3

x ± ± ±= = =

So, the tangent line will be parallel to 4 23y x= + at,

5 34 5 340.276984 3.610323 3

x x− += = − = =


Calculus I


Product and Quotient Rule

1. Use the Product Rule to find the derivative of ( ) ( )( )2 3 24 8 12f t t t t t= − − + .

Solution There isn’t much to do here other than take the derivative using the product rule.

( ) ( ) ( ) ( )( )3 2 2 2 4 3 28 1 8 12 4 3 16 20 132 24 96 12f t t t t t t t t t t t t′ = − − + + − − = − + + − Note that we multiplied everything out to get a “simpler” answer.

2. Use the Product Rule to find the derivative of ( )( )3 3 31 2y x x x−= + − .

Solution There isn’t much to do here other than take the derivative using the product rule. We’ll also need to convert the roots to fractional exponents.

13

3 321 2y x x x− = + −

The derivative is then,

1 2 2 51 3 5

3 4 43 3 3 62 2 23 2 3 2 112 1 3 32 3 2 3 3

dy x x x x x x x x x xdx

− −−− − − = − + + − − = − − − −

Note that we multiplied everything out to get a “simpler” answer.

3. Use the Product Rule to find the derivative of ( ) ( )( )2 2 31 2 3 5 8h z z z z z z= + + + − .

Solution There isn’t much to do here other than take the derivative using the product rule.


Calculus I


( ) ( ) ( ) ( ) ( )2 3 2 2

2 3 4

2 6 5 8 1 2 3 5 16 3

5 36 90 88 15

h z z z z z z z z z

z z z z

′ = + + − + + + + −

= + + + −

Note that we multiplied everything out to get a “simpler” answer.

4. Use the Quotient Rule to find the derivative of ( )26

2xg x

x=

− .

Solution There isn’t much to do here other than take the derivative using the quotient rule.

( ) ( ) ( )( ) ( )

2 2

2 2

12 2 6 1 24 62 2

x x x x xg xx x

− − − −′ = =− −

5. Use the Quotient Rule to find the derivative of ( )4

2

32 1w wR ww

+=

+ .


( ) ( )( ) ( ) ( )( ) ( )

3 2 4 5 3 2

2 22 2

3 4 2 1 3 4 4 4 6 3

2 1 2 1

w w w w w w w wR ww w

+ + − + + − +′ = =+ +

6. Use the Quotient Rule to find the derivative of ( ) 2

27 4

x xf xx x

+=

− .



Calculus I


( )( )( ) ( )( )

( )

1 121 2 22

22

7 4 7 82 27 4

x x xx x xf xx x

− − − −+ +′ =−

7. If ( )2 8f = − , ( )2 3f ′ = , ( )2 17g = and ( )2 4g′ = − determine the value of ( ) ( )2f g ′ .

Solution We know that the product rule is,

( ) ( ) ( ) ( ) ( ) ( )f g x f x g x f x g x′ ′ ′= + Now, we want to know the value of this at 2x = and so all we need to do is plug this into the derivative. Doing this gives,

( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2f g f g f g′ ′ ′= + Now, we were given values for all these quantities and so all we need to do is plug these into our “formula” above.

( ) ( ) ( )( ) ( ) ( )2 3 17 8 4 83f g ′ = + − − = 8. If ( ) ( )3f x x g x= , ( )7 2g − = , ( )7 9g′ − = − determine the value of ( )7f ′ − .

Hint : Even though we don’t know what ( )g x is we can still use the product rule to take the

derivative and then we can use the given information to get the value of ( )7f ′ − .

Solution Even though we don’t know what ( )g x is we do have a product of two functions here and so we

can use the product rule to determine the derivative of ( )f x .

( ) ( ) ( )2 33f x x g x x g x′ ′= + Now all we need to do is plug 7x = − into this and use the given information to determine the value of ( )7f ′ − .


Calculus I


( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( )2 37 3 7 7 7 7 3 49 2 343 9 3381f g g′ ′− = − − + − − = + − − =

9. Find the equation of the tangent line to ( ) ( ) ( )21 12 4f x x x= + − at 9x = .

Solution Step 1 We know that the derivative of the function will give us the slope of the tangent line so we’ll need the derivative of the function. We’ll use the product rule to get the derivative.

( ) ( ) ( )( ) ( ) ( )1

2 22 66 4 1 12 2 4 2 1 12f x x x x x x x xx

− ′ = − + + − = − − +

Step 2 Note that we didn’t bother to “simplify” the derivative (other than converting the fractional exponent back to a root) because all we really need this for is a quick evaluation. Speaking of which here are the evaluations that we’ll need for this problem. ( ) ( )( ) ( ) ( ) ( ) ( )9 37 77 2849 9 2 77 18 37 820f f ′= − = − = − − = − Step 3 Now all that we need to do is write down the equation of the tangent line.

( ) ( )( ) ( )9 9 9 2849 820 9 820 4531y f f x x y x′= + − = − − − → = − +


21 8x xf x

x−

=+

is increasing and decreasing.

Solution Step 1 We’ll first need the derivative, which will require the quotient rule, because we know that the derivative will give us the rate of change of the function. Here is the derivative.

( )( ) ( ) ( )( )

( ) ( )

2 2 2

2 22 2

1 2 1 8 16 1 2 8

1 8 1 8

x x x x x x xf xx x

− + − − − −′ = =+ +


Calculus I


Step 2 Next, we need to know where the function is not changing and so all we need to do is set the derivative equal to zero and solve. In this case it is clear that the denominator will never be zero for any real number and so the derivative will only be zero where the numerator is zero. Therefore, setting the numerator equal to zero and solving gives,

( ) ( )( )2 21 2 8 8 2 1 4 1 2 1 0x x x x x x− − = − + − = − − + = From this it is pretty easy to see that the derivative will be zero, and hence the function will not be changing, at,

1 12 4

x x= − =

Step 3 To get the answer to this problem all we need to know is where the derivative is positive (and hence the function is increasing) or negative (and hence the function is decreasing). Because the derivative is continuous we know that the only place it can change sign is where the derivative is zero. So, as we did in this section a quick number line will give us the sign of the derivative for the various intervals. Here is the number line for this problem.


1 1Increasing :2 4

1 1Decreasing : ,2 4

x

x x

− < <

− ∞ < < − < < ∞

11. Determine where ( ) ( )( )2 24 1 5V t t t= − + is increasing and decreasing.


Calculus I


Solution Step 1 We’ll first need the derivative, for which we will use the product rule, because we know that the derivative will give us the rate of change of the function. Here is the derivative.

( ) ( ) ( ) ( )( ) ( )2 2 3 22 1 5 4 10 38 20 2 19 10V t t t t t t t t t′ = − + + − = − = −

Step 2 Next, we need to know where the function is not changing and so all we need to do is set the derivative equal to zero and solve. From the factored form of the derivative it is easy to see that the derivative will be zero at,

190 1.378410

t t= = ± = ±



19 19Increasing : , 010 10

19 19Decreasing : 0,10 10

t t

t x

− ∞ < < − < <

− < < < < ∞


Calculus I


Derivatives of Trig Functions

1. Evaluate ( )

0

sin 10limz

zz→

.

Solution All we need to do is set this up to allow us to use the fact from the notes in this section.

( ) ( ) ( ) ( )0 0 0

sin 10 10sin 10 sin 10lim lim 10lim 10 1 10

10 10z z z

z z zz z z→ → →

= = = =

2. Evaluate ( )( )0

sin 12lim

sin 5α

αα→

.


( )( )

( )( )

( )( )

( )( )

( )( )

( ) ( )

0 0 0

0 0 0

sin 12 12 sin 12 sin 125 12 5lim lim limsin 5 12 5 sin 5 5 12 sin 5

sin 12 sin 1212 5 12 5lim lim lim5 12 sin 5 5 12 sin 5

12 121 15 5

α α α

α α α

α α α αα α αα α α α α α α

α αα αα α α α

→ → →

→ → →

= =

= =

= =

3. Evaluate ( )

0

cos 4 1limx

xx→

− .



Calculus I


( ) ( )( ) ( ) ( )

0 0 0

4 cos 4 1cos 4 1 cos 4 1lim lim 4lim 4 0 0

4 4x x x

xx xx x x→ → →

−− −= = = =

4. Differentiate ( ) ( ) ( )2cos 6sec 3f x x x= − + .

Solution Not much to do here other than take the derivative. ( ) ( ) ( ) ( )2sin 6sec tanf x x x x′ = − − 5. Differentiate ( ) ( ) ( )10 tan 2cotg z z z= − .

Solution Not much to do here other than take the derivative. ( ) ( ) ( )2 210sec 2cscg z z z′ = + 6. Differentiate ( ) ( ) ( )tan secf w w w= .

Solution Not much to do here other than take the derivative, which will require the product rule.

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 3 2sec sec tan sec tan sec sec tanf w w w w w w w w w′ = + = + 7. Differentiate ( ) ( )3 2 sinh t t t t= − .

Solution


Calculus I


Not much to do here other than take the derivative, which will require the product rule for the second term. You’ll need to be careful with the minus sign on the second term. You can either use a set of parenthesis around the derivative of the second term or you can think of the minus sign as part of the “first” function. We’ll think of the minus sign as part of the first function for this problem.

( ) ( ) ( )2 23 2 sin cosh t t t t t t′ = − −

8. Differentiate ( )6 4 cscy x x= + .

Solution Not much to do here other than take the derivative, which will require the product rule for the second term.

( ) ( ) ( ) ( )( ) ( ) ( ) ( )1 1

1 2 224 csc 4 csc cos 2 csc 4 csc cosy x x x x x x x x x x

− −′ = + − = −

9. Differentiate ( ) ( ) ( )1

2sin 4cosR t

t t=

− .

Solution Not much to do here other than take the derivative, which will require the quotient rule.

( ) ( ) ( ) ( )( ) ( ) ( ) ( )( )( ) ( )( )

( ) ( )( ) ( )( )2 2

0 2sin 4cos 1 2cos 4sin 2cos 4sin

2sin 4cos 2sin 4cos

t t t t t tR t

t t t t

− − + − −′ = =

− −

10. Differentiate ( ) ( )( )

tan1 cscv v

Z vv

+=

+ .

Solution Not much to do here other than take the derivative, which will require the quotient rule.


Calculus I


( )( )( ) ( )( ) ( )( ) ( ) ( )( )

( )( )( )( ) ( )( ) ( ) ( ) ( )( )

( )( )

2

2

2

2

1 sec 1 csc tan csc cot

1 csc

1 sec 1 csc csc cot tan

1 csc

v v v v v vZ v

v

v v v v v v

v

+ + − + −′ =

+

+ + + +=

+

11. Find the tangent line to ( ) ( ) ( )tan 9cosf x x x= + at x π= .

Solution Step 1 We know that the derivative of the function will give us the slope of the tangent line so we’ll need the derivative of the function. ( ) ( ) ( )2sec 9sinf x x x′ = − Step 2 Now all we need to do is evaluate the function and the derivative at the point in question. ( ) ( ) ( ) ( ) ( ) ( )2tan 9cos 9 sec 9sin 1f fπ π π π π π′= + = − = − = Step 3 Now all that we need to do is write down the equation of the tangent line.

( ) ( ) ( ) ( ) ( )9 1 9y f f x x y xπ π π π π′= + − = − + − → = − + Don’t get excited about the presence of the π in the answer. It is just a number like the 9 is and so is nothing to worry about. 12. The position of an object is given by ( ) ( )2 7coss t t= + determine all the points where the

object is not moving. Solution We know that the object will not be moving if its velocity, which is simply the derivative of the position function, is not zero. So all we need to do is take the derivative, set it equal to zero and solve.


Calculus I


( ) ( ) ( )7sin 7sin 0s t t t′ = − ⇒ − = So, for this problem the object will not be moving anywhere that sine is zero. From our recollection of the unit circle we know that will be at,

0 2 2 and 2 0, 1, 2, 3,t n n t n nπ π π π= + = = + = ± ± ± K 13. Where in the range [ ]2,7− is the function ( ) ( )4cosf x x x= − is increasing and

decreasing. Solution Step 1 We’ll first need the derivative because we know that the derivative will give us the rate of change of the function. Here is the derivative.

( ) ( )4sin 1f x x′ = − − Step 2 Next, we need to know where the function is not changing and so all we need to do is set the derivative equal to zero and solve.

( ) ( ) 14sin 1 0 sin4

x x− − = ⇒ = −

A quick calculator computation tells us that,

1 1sin 0.25274

x − = − = −

Recalling our work in the Review chapter on solving trig equations we know that a positive angle corresponding to this solution is : 2 0.2527 6.0305x π= − = . Either can be used, but we will use the positive angle. Also, from a quick check on a unit circle we can see that 0.2527 3.3943x π= + = will be a second solution. Putting all of this together and we can see that the derivative will be zero at, 6.0305 2 and 3.3943 2 0, 1, 2, 3,x n x n nπ π= + = + = ± ± ± K


Calculus I


Finally, all we need to so is plug in some n’s to determine which solutions fall in the interval we are working on, [ ]2,7− .

1: 0.2527 2.8889n x x= − = − = −

0 : 6.0305 3.39431: 12.3137

n x xn x

= = == = 9.6775x =

So, in the interval [ ]2,7− the function will stop changing at the following three points.

0.2527, 3.3943, 6.0305x = − Step 3 To get the answer to this problem all we need to know is where the derivative is positive (and hence the function is increasing) or negative (and hence the function is decreasing). Because the derivative is continuous we know that the only place it can change sign is where the derivative is zero. So, as we did in this section a quick number line will give us the sign of the derivative for the various intervals. Here is the number line for this problem.


Increasing : 2 0.2527, 3.3943 6.0305Decreasing : 0.2527 3.3943, 6.0305 7

x xx x

− ≤ < − < <− < < < ≤

Note that because we’ve only looked at what is happening in the interval [ ]2,7− we can’t say

anything about the increasing/decreasing nature of the function outside of this interval.


Calculus I


Derivatives of Exponential and Logarithm Functions 1. Differentiate ( ) 2 8xxf x = −e .

Solution Not much to do here other than take the derivative using the formulas from class.

( ) ( )2 8 ln 8xxf x′ = −e 2. Differentiate ( ) ( ) ( )34log lng t t t= − .


( ) ( )4 1

ln 3g t

t t′ = −

3. Differentiate ( ) ( )3 logwR w w= .


( ) ( ) ( ) ( )33 ln 3 log

ln 10

wwR w w

w′ = +

Recall that ( )log x is the common logarithm and so is really ( )10log x .

4. Differentiate ( )5 lnzy z z= −e .


Calculus I



( )45 lnz

zy z zz

′ = − −ee

5. Differentiate ( )1 y

yh y =−e

.


( )( )( ) ( )

( ) ( )2 2

1 1 1

1 1

y y y y

y y

y yh y− − − − +′ = =

− −

e e e e

e e

6. Differentiate ( ) ( )1 5ln

tf tt

+= .


( ) ( )

( ) ( )

( )

( )

( )2 2

1 15ln 1 5 5ln 51 5ln ln ln

t t tt t tf tt t t

+ + + + + = = =

7. Find the tangent line to ( ) 7 4x xf x = + e at 0x = .

Solution Step 1


Calculus I


We know that the derivative of the function will give us the slope of the tangent line so we’ll need the derivative of the function. ( ) ( )7 ln 7 4x xf x′ = + e Step 2 Now all we need to do is evaluate the function and the derivative at the point in question. ( ) ( ) ( )0 11 0 ln 7 4 5.9459f f ′= = + = Step 3 Now all that we need to do is write down the equation of the tangent line.

( ) ( )( ) ( )( )0 0 0 11 ln 7 4 11 5.9459y f f x x x′= + − = + + = +

8. Find the tangent line to ( ) ( ) ( )2ln logf x x x= at 2x = .

Solution Step 1 We know that the derivative of the function will give us the slope of the tangent line so we’ll need the derivative of the function.

( ) ( ) ( )( )

2log lnln 2

x xf x

x x′ = +

Step 2 Now all we need to do is evaluate the function and the derivative at the point in question.

( ) ( ) ( ) ( ) ( ) ( ) ( )( )

22

log 2 ln 22 ln 2 log 2 ln 2 2 1

2 2ln 2f f ′= = = + =


( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 ln 2 1 2 2 ln 2y f f x x x′= + − = + − = − +

9. Determine if ( ) ttV t =e

is increasing or decreasing at the following points.

(a) 4t = − (b) 0t = (c) 10t = Solution


Calculus I


(a) 4t = − We know that the derivative of the function will give us the rate of change for the function and so we’ll need that.

( )( ) ( )

( ) ( )2 2

1 1t t t t

tt t

t t tV t− − −′ = = =

e e e eee e

Now, all we need to do is evaluate the derivative at the point in question. So,

( ) 454 272.991 0V −′ − = = >

e

( )4 0V ′ − > and so the function must be increasing at 4t = − .

(b) 0t = We found the derivative of the function in the first part so here all we need to do is the evaluation.

( ) 010 1 0V ′ = = >e

( )0 0V ′ > and so the function must be increasing at 0t = .

(c) 10t = We found the derivative of the function in the first part so here all we need to do is the evaluation.

( ) 10910 0.0004086 0V −′ = = − <

e

( )10 0V ′ < and so the function must be decreasing at 10t = .

10. Determine if ( ) ( ) ( )6 lnG z z z= − is increasing or decreasing at the following points.

(a) 1z = (b) 5z = (c) 20z = Solution (a) 1z = We know that the derivative of the function will give us the rate of change for the function and so we’ll need that.


Calculus I


( ) ( ) 6ln zG z zz−′ = +

Now, all we need to do is evaluate the derivative at the point in question. So, ( ) ( )1 ln 1 5 5 0G′ = − = − <

( )1 0G′ < and so the function must be decreasing at 1z = .

(b) 5z = We found the derivative of the function in the first part so here all we need to do is the evaluation.

( ) ( ) 15 ln 5 1.40944 05

G′ = − = >

( )5 0G′ > and so the function must be increasing at 5z = .

(c) 20z = We found the derivative of the function in the first part so here all we need to do is the evaluation.

( ) ( ) 720 ln 20 3.6957310

G′ = + =

( )20 0G′ > and so the function must be increasing at 20z = .

Derivatives of Inverse Trig Functions 1. Differentiate ( ) ( ) ( )12cos 6cosT z z z−= + .


( ) ( )2

62sin1

T z zz

′ = − −−


Calculus I


2. Differentiate ( ) ( ) ( )1 1csc 4cotg t t t− −= − .


( ) 22

1 411

g ttt t

′ = − ++−

3. Differentiate ( )6 15 secy x x−= − .


5

2

1301

dy xdx x x

= −−

4. Differentiate ( ) ( ) ( )2 1sin tanf w w w w−= + .


( ) ( ) ( )2

12cos 2 tan

1wf w w w w

w−′ = + +

+

5. Differentiate ( ) ( )1sin1

xh x

x

−

=+

.



Calculus I


( )( )

( )( )

( )

12 12

2 22

1 sin1 1 sin1

1 1 1

x xx x xxh x

x x x

−−

+ −+ − −−′ = =

+ − +

Derivatives of Hyperbolic Functions 1. Differentiate ( ) ( ) ( ) ( )sinh 2cosh sechf x x x x= + − .


( ) ( ) ( ) ( ) ( )cosh 2sinh sech tanhf x x x x x′ = + + 2. Differentiate ( ) ( ) ( )2tan cschR t t t t= + .


( ) ( ) ( ) ( ) ( )2 2sec 2 csch csch cothR t t t t t t t′ = + −

3. Differentiate ( ) ( )1

tanhzg z

z+

= .



Calculus I


( ) ( ) ( ) ( )( )

2

2

tanh 1 sechtanh

z z zg z

z− +

′ =

Chain Rule

1. Differentiate ( ) ( )426 7f x x x= + .

Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution For this problem the outside function is (hopefully) clearly the exponent of 4 on the parenthesis while the inside function is the polynomial that is being raised to the power. The derivative is then,

( ) ( ) ( ) ( )( )3 32 24 6 7 12 7 4 12 7 6 7f x x x x x x x′ = + + = + +

2. Differentiate ( ) ( ) 224 3 2g t t t−

= − + .

Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution For this problem the outside function is (hopefully) clearly the exponent of -2 on the parenthesis while the inside function is the polynomial that is being raised to the power. The derivative is then,

( ) ( ) ( ) ( ) ( )3 32 22 4 3 2 8 3 2 8 3 4 3 2g t t t t t t t− −

′ = − − + − = − − − +


Calculus I


3. Differentiate 3 1 8y z= − . Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution For this problem, after converting the root to a fractional exponent, the outside function is

(hopefully) clearly the exponent of 13

while the inside function is the polynomial that is being

raised to the power (or the polynomial inside the root – depending upon how you want to think about it). The derivative is then,

( ) ( ) ( ) ( )1 2 23 3 3

1 81 8 1 8 8 1 83 3

dyy z z zdz

− −= − ⇒ = − − = − −

4. Differentiate ( ) ( )csc 7R w w= .

Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution For this problem the outside function is (hopefully) clearly the trig function and the inside function is the stuff inside of the trig function. The derivative is then,

( ) ( ) ( )7csc 7 cot 7R w w w′ = − In dealing with functions like cosecant (or secant for that matter) be careful to make sure that the inside function gets substituted into both terms of the derivative of the outside function. One of the more common mistakes with this kind of problem is to only substitute the 7w into only the cosecant or only the cotangent instead of both as it should be. 5. Differentiate ( ) ( )( )2sin 3 tanG x x x= + .


Calculus I


Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution For this problem the outside function is (hopefully) clearly the sine function and the inside function is the stuff inside of the trig function. The derivative is then,

( ) ( )( ) ( )( )22 3 sec cos 3 tanG x x x x′ = + +

6. Differentiate ( ) ( )tan 4 10h u u= + .

Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution For this problem the outside function is (hopefully) clearly the trig function and the inside function is the stuff inside of the trig function. The derivative is then,

( ) ( )210sec 4 10h u u′ = +

7. Differentiate ( ) 745 t tf t += + e .

Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution Note that we only need to use the Chain Rule on the second term as we can differentiate the first term without the Chain Rule. Now, recall that for exponential functions outside function is the exponential function itself and the inside function is the exponent. The derivative is then,

( ) ( ) 76 44 7 t tf t t +′ = + e


Calculus I


8. Differentiate ( ) ( )1 cos xg x −= e .

Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution For exponential functions remember that the outside function is the exponential function itself and the inside function is the exponent. The derivative is then,

( ) ( ) ( )1 cossin xg x x −′ = e 9. Differentiate ( ) 1 62 zH z −= .

Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution For exponential functions remember that the outside function is the exponential function itself and the inside function is the exponent. The derivative is then,

( ) ( ) ( )1 66 2 ln 2zH z −′ = −

10. Differentiate ( ) ( )1tan 3 1u t t−= − .

Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution For this problem the outside function is (hopefully) clearly the inverse tangent and the inside function is the stuff inside of the inverse tangent. The derivative is then,


Calculus I


( )( )2

33 1 1

u tt

′ =− +

11. Differentiate ( ) ( )2 3ln 1 5F y y y= − + .

Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution For this problem the outside function is (hopefully) clearly the logarithm and the inside function is the stuff inside of the logarithm. The derivative is then,

( ) ( )2

22 3 2 3

1 10 310 31 5 1 5

y yF y y yy y y y

− += − + =

− + − +

With logarithm problems remember that after differentiating the logarithm (i.e. the outside function) you need to substitute the inside function into the derivative. So, instead of getting just,

1y

we get the following (i.e. we plugged the inside function into the derivative),

2 3

11 5y y− +

Then, we can’t forget of course to multiply by the derivative of the inside function. 12. Differentiate ( ) ( ) ( )( )ln sin cotV x x x= − .

Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution For this problem the outside function is (hopefully) clearly the logarithm and the inside function is the stuff inside of the logarithm. The derivative is then,


Calculus I


( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( )

22 cos csc1 cos csc

sin cot sin cotx x

V x x xx x x x

+= + =

− −

With logarithm problems remember that after differentiating the logarithm (i.e. the outside function) you need to substitute the inside function into the derivative. So, instead of getting just,

1x

we get the following (i.e. we plugged the inside function into the derivative),

( ) ( )

1sin cotx x−

Then, we can’t forget of course to multiply by the derivative of the inside function.

13. Differentiate ( ) ( ) ( )6 6sin sinh z z z= + .

Hint : Don’t get too locked into problems only requiring a single use of the Chain Rule. Sometimes separate terms will require different applications of the Chain Rule, or maybe only one of the terms will require the Chain Rule. Solution For this problem each term will require a separate application of the Chain Rule and don’t forget that,

( ) ( ) 66sin sinz z= So, in the first term the outside function is the sine function, while the sine function is the inside function in the second term. The derivative is then,

( ) ( ) ( ) ( )5 6 56 cos 6sin cosh z z z z z′ = +

14. Differentiate ( ) 7 wS w w −= + e .


Calculus I


Hint : Don’t get too locked into problems only requiring a single use of the Chain Rule. Sometimes separate terms will require different applications of the Chain Rule, or maybe only one of the terms will require the Chain Rule. Solution For this problem each term will require a separate application of the Chain Rule and make sure you are careful with parenthesis in dealing with the root in the first term. The derivative is then,

( ) ( ) ( ) ( ) ( ) ( )1 1 12 2 2

1 77 7 7 72 2

w w wS w w S w w w− −− − −′= + ⇒ = − = −e e e

15. Differentiate ( ) ( )7 23 sin 6g z z z= − + .

Hint : Don’t get too locked into problems only requiring a single use of the Chain Rule. Sometimes separate terms will require different applications of the Chain Rule, or maybe only one of the terms will require the Chain Rule. Solution For this problem the first term requires no Chain Rule and the second term will require the Chain Rule. The derivative is then,

( ) ( )6 221 2 cos 6g z z z z′ = − +

16. Differentiate ( ) ( )( ) ( )104ln sin 3f x x x x= − − .

Hint : Don’t get too locked into problems only requiring a single use of the Chain Rule. Sometimes separate terms will require different applications of the Chain Rule, or maybe only one of the terms will require the Chain Rule. Solution For this problem each term will require a separate application of the Chain Rule. The derivative is then,


Calculus I


( ) ( )( ) ( )( ) ( ) ( )( )9 93 4 3 4cos

10 4 3 3 cot 10 4 3 3sin

xf x x x x x x x x

x′ = − − − = − − −

17. Differentiate ( ) 6 25h t t t t= − .

Hint : Don’t forget the Product and Quotient Rule. Sometimes, in the process of doing the Product or Quotient Rule you’ll need to use the Chain Rule when differentiating one or both of the terms in the product or quotient. Solution For this problem we’ll need to do the Product Rule to start off the derivative. In the process we’ll need to use the Chain Rule when we differentiate the second term. The derivative is then,

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )( )

16 2 2

1 1 1 15 2 6 2 5 2 6 22 2 2 2

5

1 16 5 5 10 1 6 5 10 1 52 2

h t t t t

h t t t t t t t t t t t t t t t− −

= −

′ = − + − − = − + − −

18. Differentiate ( ) ( )2 5lnq t t t= .

Hint : Don’t forget the Product and Quotient Rule. Sometimes, in the process of doing the Product or Quotient Rule you’ll need to use the Chain Rule when differentiating one or both of the terms in the product or quotient. Solution For this problem we’ll need to do the Product Rule to start off the derivative. In the process we’ll need to use the Chain Rule when we differentiate the second term. The derivative is then,


Calculus I


( ) ( ) ( )4

5 2 55

52 ln 2 ln 5tq t t t t t t tt

′ = + = +

19. Differentiate ( ) ( ) ( )cos 3 sec 1g w w w= − .

Hint : Don’t forget the Product and Quotient Rule. Sometimes, in the process of doing the Product or Quotient Rule you’ll need to use the Chain Rule when differentiating one or both of the terms in the product or quotient. Solution For this problem we’ll need to do the Product Rule to start off the derivative. In the process we’ll need to use the Chain Rule when we differentiate each term. The derivative is then,

( ) ( ) ( ) ( ) ( ) ( ) ( )( )

( ) ( ) ( ) ( ) ( )sin 3 3 sec 1 cos 3 sec 1 tan 1 1

3sin 3 sec 1 cos 3 sec 1 tan 1

g w w w w w w

w w w w w

′ = − − + − − −

= − − − − −

20. Differentiate ( )

2

sin 31

ty

t=

+ .

Hint : Don’t forget the Product and Quotient Rule. Sometimes, in the process of doing the Product or Quotient Rule you’ll need to use the Chain Rule when differentiating one or both of the terms in the product or quotient. Solution For this problem we’ll need to do the Quotient Rule to start off the derivative. In the process we’ll need to use the Chain Rule when we differentiate the numerator. The derivative is then,

( )( ) ( )( )

( )( )( ) ( )

( )

2 2

2 22 2

3cos 3 1 sin 3 2 3cos 3 1 2 sin 3

1 1

t t t t t t t tdydt t t

+ − + −= =

+ +


Calculus I


21. Differentiate ( ) ( )21

tan 12

xK x

x x

−+=

+e

.

Hint : Don’t forget the Product and Quotient Rule. Sometimes, in the process of doing the Product or Quotient Rule you’ll need to use the Chain Rule when differentiating one or both of the terms in the product or quotient. Solution For this problem we’ll need to do the Quotient Rule to start off the derivative. In the process we’ll need to use the Chain Rule when we differentiate both the numerator and the denominator. The derivative is then,

( )( )( ) ( ) ( )( )

( )( )

2 2 2

2

2 tan 12 1 1 12sec 12

tan 12

x xx x xK x

x x

− −− + − + +′ =

+

e e

22. Differentiate ( ) ( )2cos xf x x= e .

Hint : Don’t forget the Product and Quotient Rule. Sometimes, in the process of using the Chain Rule, you’ll also need the Product and/or Quotient Rule. Solution For this problem we’ll start off using the Chain Rule, however when we differentiate the inside function we’ll need to do the Product Rule. The derivative is then,

( ) ( ) ( )2 22 sinx x xf x x x x′ = − +e e e

23. Differentiate ( )5 tan 4z x x= + .


Calculus I


Hint : Sometimes the Chain Rule will need to be done multiple times before we finish taking the derivative. Step 1 This problem will require multiple uses of the Chain Rule and so we’ll step though the derivative process to make each use clear. Here is the first step of the derivative and we’ll need to use the Chain Rule in this step.

( )( )

( )( ) ( )( )

12

12

5 tan 4

1 5 tan 4 5 tan 42

z x x

dz dx x x xdx dx

−

= +

= + +

Step 2 In this step we can see that we’ll need to use the Chain Rule on the second term. The derivative is then,

( )( ) ( )( )1

221 5 tan 4 5 4sec 42

dz x x xdx

−= + +

In this step we were using the Chain Rule on the second term and so when multiplying by the derivative of the inside function we only multiply the second term by the derivative of the inside function and not both terms.

24. Differentiate ( ) ( )( )36 sin 2tf t t−= + −e .

Hint : Sometimes the Chain Rule will need to be done multiple times before we finish taking the derivative. Step 1 This problem will require multiple uses of the Chain Rule and so we’ll step though the derivative process to make each use clear.


Calculus I


Here is the first step of the derivative and we’ll need to use the Chain Rule in this step.

( ) ( )( ) ( )( )26 63 sin 2 sin 2t tdf t t tdt

− −′ = + − + −e e

Step 2 In this step we can see that we’ll need to use the Chain Rule on each of the terms. The derivative is then,

( ) ( )( ) ( )( )26 63 sin 2 6 cos 2t tf t t t− −′ = + − − − −e e

25. Differentiate ( ) ( ) ( )( )102 1ln 1 tan 6g x x x−= + − .

Hint : Sometimes the Chain Rule will need to be done multiple times before we finish taking the derivative. Step 1 This problem will require multiple uses of the Chain Rule and so we’ll step though the derivative process to make each use clear. Here is the first step of the derivative and we’ll need to use the Chain Rule in this step.

( ) ( ) ( )( ) ( ) ( )( )92 1 2 110 ln 1 tan 6 ln 1 tan 6dg x x x x xdx

− −′ = + − + −

Step 2 In this step we can see that we’ll need to use the Chain Rule on each of the terms. The derivative is then,

( ) ( ) ( )( )92 12 2

2 610 ln 1 tan 61 36 1

xg x x xx x

− ′ = + − − + +


Calculus I


26. Differentiate ( ) ( )4 2tan 1h z z= + .

Hint : Sometimes the Chain Rule will need to be done multiple times before we finish taking the derivative. Step 1 This problem will require multiple uses of the Chain Rule and so we’ll step though the derivative process to make each use clear. Also, recall that,

( ) ( ) 44tan tanx x=


( ) ( ) ( )3 2 24 tan 1 tan 1dh z z zdz

′ = + + Step 2 As we can see the derivative from the previous step will also require the Chain Rule. The derivative is then,

( ) ( ) ( ) ( ) ( ) ( )3 2 2 2 3 2 2 24 tan 1 sec 1 2 8 tan 1 sec 1h z z z z z z z′ = + + = + +

27. Differentiate ( ) ( )( ) 123 12 sin 3f x x x

−= + .

Hint : Sometimes the Chain Rule will need to be done multiple times before we finish taking the derivative. Step 1 This problem will require multiple uses of the Chain Rule and so we’ll step though the derivative process to make each use clear.


Calculus I



( ) ( )( ) ( ) ( )12

2 23 312 sin 3 12 sin 3df x x x x xdx

− ′ = − + +

Step 2 As we can see the derivative from the previous step will also require the Chain Rule on each of the terms. The derivative from this step is,

( ) ( )( ) ( ) ( ) ( ) ( )( )22

23 3112 sin 3 12 12 2sin 3 sin 33

df x x x x x xdx

− − ′ = − + +

Step 3 The second term will again use the Chain Rule as we can see. The derivative is then,

( ) ( )( ) ( ) ( ) ( )22

23 312 sin 3 4 12 6sin 3 cos 3f x x x x x x− − ′ = − + +

28. Find the tangent line to ( ) 24 2 6 xf x x −= − e at 2x = .

Solution Step 1 We know that the derivative of the function will give us the slope of the tangent line so we’ll need the derivative of the function. Differentiating each term will require the Chain Rule as well.

( ) ( )

( ) ( ) ( ) ( ) ( )

122

1 12 2 22 2

4 2 6

1 44 2 2 6 1 4 2 6 62 2

x

x x x

f x x

f x x xx

−

− −− − −

= −

′ = − − = + = +

e

e e e

Step 2 Now all we need to do is evaluate the function and the derivative at the point in question.

( ) ( ) ( )0 042 4 2 6 2 2 6 82

f f ′= − = = + =e e


Calculus I



( ) ( ) ( ) ( )2 2 2 2 8 2 8 14y f f x x y x′= + − = + − → = −

29. Determine where ( ) ( )34 2 8V z z z= − is increasing and decreasing.

Solution Step 1 We’ll first need the derivative because we know that the derivative will give us the rate of change of the function. Here is the derivative.

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

3 23 4

2 23 3

4 2 8 3 2 8 2

2 2 8 2 2 8 3 2 2 8 7 16

V z z z z z

z z z z z z z

′ = − + −

= − − + = − −

Note that we factored the derivative to help with the next step. In general we don’t need to do this. Step 2 Next, we need to know where the function is not changing and so all we need to do is set the derivative equal to zero and solve. In this case it’s pretty easy to spot where the derivative will be zero.

( ) ( )23 162 2 8 7 16 0 0, 4, 2.28577

z z z z z z− − = ⇒ = = = =



Calculus I



16Increasing : 0, 4, 47

16Decreasing : 07

x x x

x

− ∞ < < < < < < ∞

< <

30. The position of an object is given by ( ) ( )sin 3 2 4s t t t= − + . Determine where in the

interval [ ]0,3 the object is moving to the right and moving to the left.

Solution Step 1 We’ll first need the derivative because we know that the derivative will give us the velocity of the object. Here is the derivative.

( ) ( )3cos 3 2s t t′ = − Step 2 Next, we need to know where the object stops moving and so all we need to do is set the derivative equal to zero and solve.

( ) ( ) 23cos 3 2 0 cos 33

t t− = ⇒ =

A quick calculator computation tells us that,

1 23 cos 0.84113

t − = =


Calculus I


Recalling our work in the Review chapter and a quick check on a unit circle we can see that either 3 0.8411t = − or 3 2 0.8411 5.4421t π= − = can be used for the second angle. Note that either will work, but we’ll use the second simply because it is the positive angle. Putting all of this together and dividing by 3 we can see that the derivative will be zero at,

3 0.8411 2 and 3 5.4421 2 0, 1, 2, 3,

2 20.2804 and 1.8140 0, 1, 2, 3,3 3

t n t n nn nt t n

π ππ π

= + = + = ± ± ±

= + = + = ± ± ±

K

K

Finally, all we need to so is plug in some n’s to determine which solutions fall in the interval we are working on, [ ]0,6 .

0 : 0.2804 1.81401: 2.3748 3.9084

n t tn t t

= = == = =

So, in the interval [ ]0,6 , the object stops moving at the following three points.

0.2804, 1.8140, 2.3748t = Step 3 To get the answer to this problem all we need to know is where the derivative is positive (and hence the object is moving to the right) or negative (and hence the object is moving to the left). Because the derivative is continuous we know that the only place it can change sign is where the derivative is zero. So, as we did in this section a quick number line will give us the sign of the derivative for the various intervals. Here is the number line for this problem.

From this we get the following moving right/moving left information.

Moving Right : 0 0.2804, 1.8140 2.3748Moving Left : 0.2804 1.8140, 2.3748 3

t xx x

≤ < < << < < ≤


Calculus I


Note that because we’ve only looked at what is happening in the interval [ ]0,3 we can’t say

anything about the moving right/moving left behavior of the object outside of this interval. 31. Determine where ( ) 2 5 tA t t −= e is increasing and decreasing.

Solution Step 1 We’ll first need the derivative because we know that the derivative will give us the rate of change of the function. Here is the derivative.

( ) ( )5 2 5 52 2t t tA t t t t t− − −′ = − = −e e e Note that we factored the derivative to help with the next step. In general we don’t need to do this. Step 2 Next, we need to know where the function is not changing and so all we need to do is set the derivative equal to zero and solve. In this case it’s pretty easy to spot where the derivative will be zero. ( )5 2 0 0, 2tt t t t− − = ⇒ = =e Step 3 To get the answer to this problem all we need to know is where the derivative is positive (and hence the function is increasing) or negative (and hence the function is decreasing). Because the derivative is continuous we know that the only place it can change sign is where the derivative is zero. So, as we did in this section a quick number line will give us the sign of the derivative for the various intervals. Here is the number line for this problem.



Calculus I


Increasing :0 2Decreasing : 0, 2

tt t

< <− ∞ < < < < ∞

32. Determine where in the interval [ ]1, 20− the function ( ) ( )4 3ln 20 100f x x x= − − is

increasing and decreasing. Solution Step 1 We’ll first need the derivative because we know that the derivative will give us the rate of change of the function. Here is the derivative.

( ) ( )23 2

4 3 4 3

4 154 6020 100 20 100

x xx xf xx x x x

−−= =

− − − −

Note that we factored the numerator to help with the next step. In general we don’t need to do this. Step 2 Next, we need to know where the function is not changing and so all we need to do is set the derivative equal to zero and solve.

( ) ( )

22

4 3

4 150 4 15 0 0, 15

20 100x x

x x x xx x

−= → − = ⇒ = =

− −

Note, that in general, we also need to be concerned with where the derivative is not defined as well. Functions can (and often do) change sign where they are not defined. In this case however we’ve restricted the interval down to a range where the function exists and is continuous on the given interval and so this is something we need to worry about for this problem. In the next Chapter we will start also looking at what happens if the derivative is also not defined at particular points. Note as well that we really should at this point step back and recall that we are working on the interval [ ]1, 20− . We are only interested in what is happening on this interval and so we should

make sure that the points found above are inside the interval. In this case both points are inside the interval we are interested in. However, if any of them had been outside of the interval we would have needed to exclude that point.


Calculus I




Increasing : 1 0, 0 15Decreasing : 15 20

x xx

− ≤ < < << ≤

Note that because we’ve only looked at what is happening in the interval [ ]1, 20− we can’t say

anything about the increasing/decreasing nature of the function outside of this interval.

Implicit Differentiation

1. For 3 1xy

= do each of the following.


(a) Find y′ by solving the equation for y and differentiating directly. Step 1 First we just need to solve the equation for y.


Calculus I


1

3 3y x y x= ⇒ = Step 2 Now differentiate with respect to x.

2

1 33y x

−′ =

(b) Find y′ by implicit differentiation.

Hint : Don’t forget that y is really ( )y x and so we’ll need to use the Chain Rule when taking the

derivative of terms involving y! Also, don’t forget that because y is really ( )y x we may well

have a Product and/or a Quotient Rule buried in the problem. Step 1 First, we just need to take the derivative of everything with respect to x and we’ll need to recall that y is really ( )y x and so we’ll need to use the Chain Rule when taking the derivative of terms

involving y. Also, prior to taking the derivative a little rewrite might make this a little easier. 3 1x y− = Now take the derivative and don’t forget that we actually have a product of functions of x here and so we’ll need to use the Product Rule when differentiating the left side. 3 43 0y x y y− − ′− = Step 2 Finally all we need to do is solve this for y′ .

3

43 3y yyxy x

−

−′ = =

(c) Check that the derivatives in (a) and (b) are the same. Hint : To show they are the same all we need is to plug the formula for y (which we already have….) into the derivative we found in (b) and, potentially with a little work, show that we get the same derivative as we got in (a). From (a) we have a formula for y written explicitly as a function of x so plug that into the derivative we found in (b) and, with a little simplification/work, show that we get the same derivative as we got in (a).

123

1 333 3

y xy xx x

−′ = = =


Calculus I


So, we got the same derivative as we should. 2. For 2 3 4x y+ = do each of the following.

(d) Find y′ by solving the equation for y and differentiating directly. (e) Find y′ by implicit differentiation. (f) Check that the derivatives in (a) and (b) are the same.


( )1

3 2 2 34 4y x y x= − ⇒ = − Step 2 Now differentiate with respect to x.

( )2

22 33 4y x x

−′ = − −

(b) Find y′ by implicit differentiation.


derivative of terms involving y! Step 1 First, we just need to take the derivative of everything with respect to x and we’ll need to recall that y is really ( )y x and so we’ll need to use the Chain Rule when taking the derivative of terms

involving y. Taking the derivative gives, 22 3 0x y y′+ = Step 2 Finally all we need to do is solve this for y′ .

2

23

xyy

′ = −

(c) Check that the derivatives in (a) and (b) are the same. Hint : To show they are the same all we need is to plug the formula for y (which we already have….) into the derivative we found in (b) and, potentially with a little work, show that we get the same derivative as we got in (a).


Calculus I


From (a) we have a formula for y written explicitly as a function of x so plug that into the derivative we found in (b) and, with a little simplification/work, show that we get the same derivative as we got in (a).

( )

( )2

22 3322

2 3

2 2 43 3 4

x xy x xy x

−′ = − = − = − −

−

So, we got the same derivative as we should. 3. For 2 2 2x y+ = do each of the following.

(g) Find y′ by solving the equation for y and differentiating directly. (h) Find y′ by implicit differentiation. (i) Check that the derivatives in (a) and (b) are the same.


( )1

2 2 2 22 2y x y x= − ⇒ = ± −

Note that because we have no restriction on y (i.e. we don’t know if y is positive or negative) we really do need to have the “ ± ” there and that does lead to issues when taking the derivative. Hint : Two formulas for y and so two derivatives. Step 2 Now, because there are two formulas for y we will also have two formulas for the derivative, one for each formula for y. The derivatives are then,

( ) ( ) ( )

( ) ( ) ( )

1 12 22 2

1 12 22 2

2 2 0

2 2 0

y x y x x y

y x y x x y

−

−

′= − ⇒ = − − >

′= − − ⇒ = − <

As noted above the first derivative will hold for 0y > while the second will hold for 0y < and we can use either for 0y = as the plus/minus won’t affect that case. (b) Find y′ by implicit differentiation.


Calculus I




involving y. Taking the derivative gives, 2 2 0x y y′+ = Step 2 Finally all we need to do is solve this for y′ .

xyy

′ = −

(c) Check that the derivatives in (a) and (b) are the same. Hint : To show they are the same all we need is to plug the formula for y (which we already have….) into the derivative we found in (b) and, potentially with a little work, show that we get the same derivative as we got in (a). Again, two formulas for y so two derivatives… From (a) we have a formula for y written explicitly as a function of x so plug that into the derivative we found in (b) and, with a little simplification/work, show that we get the same derivative as we got in (a). Also, because we have two formulas for y we will have two formulas for the derivative. First, if 0y > we will have,

( )( )

( )1 1

2 22 21

2 2

2 22

x xy x y x xy x

−′= − ⇒ = − = − = − −

−

Next, if 0y < we will have,

( )( )

( )1 1

2 22 21

2 2

2 22

x xy x y x xy x

−′= − − ⇒ = − = − = −

− −

So, in both cases, we got the same derivative as we should. 4. Find y′ by implicit differentiation for 3 2 62 4y x y x+ − = .


Calculus I




involving y. Differentiating with respect to x gives, 2 56 8 6y y x y x′ ′+ − = Step 2 Finally all we need to do is solve this for y′ .

( )5

2 52

6 86 1 6 86 1x xy y x x yy

−′ ′− = − ⇒ =−

5. Find y′ by implicit differentiation for ( )2 47 sin 3 12y x y+ = − .



involving y. Differentiating with respect to x gives, ( ) 314 3cos 3 4y y x y y′ ′+ = − Step 2 Finally all we need to do is solve this for y′ .

( ) ( ) ( )33

3cos 314 4 3cos 3

14 4x

y y y x yy y

−′ ′+ = − ⇒ =

+

6. Find y′ by implicit differentiation for ( )sinx y x− =e .


derivative of terms involving y! Step 1


Calculus I


First, we just need to take the derivative of everything with respect to x and we’ll need to recall that y is really ( )y x and so we’ll need to use the Chain Rule when taking the derivative of terms

involving y. Differentiating with respect to x gives, ( )cos 1x y y′+ =e Don’t forget the y′ on the cosine after differentiating. Again, y is really ( )y x and so when

differentiating ( )sin y we really differentiating ( )sin y x and so we are differentiating using

the Chain Rule! Step 2 Finally all we need to do is solve this for y′ .

( ) ( ) ( )1 1 sec

cos

xxy y

y−′ = = −

e e

7. Find y′ by implicit differentiation for 2 7 5 34 2 4x y x x y− = + . Hint : Don’t forget that y is really ( )y x and so we’ll need to use the Chain Rule when taking the



involving y. This also means that the first term on the left side is really a product of functions of x and hence we will need to use the Product Rule when differentiating that term. Differentiating with respect to x gives, 7 2 6 4 28 28 2 5 12xy x y y x y y′ ′+ − = + Step 2 Finally all we need to do is solve this for y′ .

( )7 4

7 4 2 2 62 2 6

8 5 28 5 2 12 2812 28

xy xxy x y x y y yy x y

− −′ ′− − = − ⇒ =−

8. Find y′ by implicit differentiation for ( ) 22cos 2 1yx y x+ + =e .


Calculus I





involving y. This also means that the second term on the left side is really a product of functions of x and hence we will need to use the Product Rule when differentiating that term. Differentiating with respect to x gives,

( ) ( ) 2 222 2 sin 2 2 0y yx y x y y y x′ ′− + + + + =e e Step 2 Finally all we need to do is solve this for y′ (with some potentially messy algebra).

( ) ( )( )( ) ( )

( )( )

2 2

2 2

2

2

2 2

2 2

2

2

2 sin 2 2 sin 2 2 0

2 2sin 2 0 2 sin 2

2 sin 2

2 2sin 2

y y

y y

y

y

x x y y x y y y x

yx x y y x x y

x x yy

yx x y

′ ′− + − + + + =

′− + = + + −

+ −′ =

− +

e e

e e

e

e

9. Find y′ by implicit differentiation for ( )2 4 2tan 3x y x y= + .




involving y. This also means that the when doing Chain Rule on the first tangent on the left side we will need to do Product Rule when differentiating the “inside term”. Differentiating with respect to x gives,

( ) ( )4 2 3 2 2 42 4 sec 3 2x y x y y x y y y′ ′+ = + Step 2 Finally all we need to do is solve this for y′ (with some potentially messy algebra).


Calculus I


( ) ( )( )( ) ( )

( )( )

4 2 2 4 2 3 2 2 4

2 3 2 2 4 4 2 2 4

4 2 2 4

2 3 2 2 4

2 sec 4 sec 3 2

4 sec 2 3 2 sec

3 2 sec

4 sec 2

x y x y x y y x y y y

x y x y y y x y x y

x y x yy

x y x y y

′ ′+ = +

′− = −

−′ =

−

10. Find the equation of then tangent line to 4 2 3x y+ = at ( )1, 2− .

Hint : We know how to compute the slope of tangent lines and with implicit differentiation that shouldn’t be too hard at this point. Step 1 The first thing to do is use implicit differentiation to find y′ for this function.

3

3 24 2 0 xx y y yy

′ ′+ = ⇒ = −

Step 2 Evaluating the derivative at the point in question to get the slope of the tangent line gives,

1, 2

2 22x y

m y= =−

′= = − =−

Step 3 Now, we just need to write down the equation of the tangent line.

( ) ( ) ( ) ( )2 2 1 2 1 2 2 2y x y x x− − = − ⇒ = − − = −

11. Find the equation of then tangent line to 2 2 23xy y x= +e at ( )0,3 .

Hint : We know how to compute the slope of tangent lines and with implicit differentiation that shouldn’t be too hard at this point. Step 1 The first thing to do is use implicit differentiation to find y′ for this function.

2 2

2 2 22

2 22 2 3 22 3

xx x

xx yyy y y x y

y−′ ′ ′+ = + ⇒ =

−ee e

e

Step 2 Evaluating the derivative at the point in question to get the slope of the tangent line gives,


Calculus I


0, 3

18 63x y

m y= =

−′= = = −

Step 3 Now, we just need to write down the equation of the tangent line.

( )3 6 0 6 3y x y x− = − − ⇒ = − + 12. Assume that ( )x x t= , ( )y y t= and ( )z z t= and differentiate 2 3 4 1x y z− + = with

respect to t. Hint : This is just implicit differentiation like we’ve been doing to this point. The only difference is that now all the functions are functions of some fourth variable, t. Outside of that there is nothing different between this and the previous problems. Solution Differentiating with respect to t gives,

2 32 3 4 0x x y y z z′ ′ ′− + =

Note that because we were not asked to give the formula for a specific derivative we don’t need to go any farther. We could however, if asked, solved this for any of the three derivatives that are present.

13. Assume that ( )x x t= , ( )y y t= and ( )z z t= and differentiate ( ) ( )2 3cos sin 4x y y z= +

with respect to t. Hint : This is just implicit differentiation like we’ve been doing to this point. The only difference is that now all the functions are functions of some fourth variable, t. Outside of that there is nothing different between this and the previous problems. Solution Differentiating with respect to t gives,

( ) ( ) ( ) ( )2 2 32 cos sin 3 4 cos 4x x y x y y y y z y z′ ′ ′ ′− = + +

Note that because we were not asked to give the formula for a specific derivative we don’t need to go any farther. We could however, if asked, solved this for any of the three derivatives that are present.


Calculus I


Related Rates 1. In the following assume that x and y are both functions of t. Given 2x = − , 1y = and 4x′ = − determine y′ for the following equation.

2 2 3 4 46 2 yy x x −+ = − e Hint : This is just like the problems worked in the section notes. The only difference is that you’ve been given the equation and all the needed information and so you won’t have to worry about finding that. Step 1 The first thing that we need to do here is use implicit differentiation to differentiate the equation with respect to t. 2 4 4 3 4 412 2 3 4y yy y x x x x x y− −′ ′ ′ ′+ = − +e e Step 2 All we need to do now is plug in the given information and solve for y′ .

81112 16 48 32y y y′ ′ ′+ = − ⇒ =

2. In the following assume that x, y and z are all functions of t. Given 4x = , 2y = − , 1z = ,

9x′ = and 3y′ = − determine z′ for the following equation.

( ) 3 2 2 21 5 3x y z y z x− + = + − Hint : This is just like the problems worked in the section notes. The only difference is that you’ve been given the equation and all the needed information and so you won’t have to worry about finding that. Step 1 The first thing that we need to do here is use implicit differentiation to differentiate the equation with respect to t. ( ) 2 2 21 15 2 2 2x y x y z z y y z y z z x x′ ′ ′ ′ ′ ′− − + = + + Step 2 All we need to do now is plug in the given information and solve for z′ .


Calculus I


45727 12 15 12 8 72z z z′ ′ ′+ + = + + ⇒ =

3. For a certain rectangle the length of one side is always three times the length of the other side.

(a) If the shorter side is decreasing at a rate of 2 inches/minute at what rate is the longer side decreasing?

(b) At what rate is the enclosed area decreasing when the shorter side is 6 inches long and is decreasing at a rate of 2 inches/minute?

Hint : The equation needed here is a really simple equation. In fact, so simple it might be easy to miss… (a) If the shorter side is decreasing at a rate of 2 inches/minute at what rate is the longer side decreasing? Step 1 Let’s call the shorter side x and the longer side y. We know that 2x′ = − and want to find y′ . Now all we need is an equation that relates these two quantities and from the problem statement we know the longer side is three times shorter side and so the equation is, 3y x= Step 2 Next step is to simply differentiate the equation with respect to t. 3y x′ ′=

Step 3

Finally, plug in the known quantity and solve for what we want : 6y′ = −

Hint : Once we have the equation for the area we can either simplify the equation as we did in this section or we can use the result from the previous step and the equation directly. (b) At what rate is the enclosed area decreasing when the shorter side is 6 inches long and is decreasing at a rate of 2 inches/minute? Step 1 Again we’ll call the shorter side x and the longer side y as with the last part. We know that 6x = ,

2x′ = − and want to find A′ . The equation we’ll need is just the area formula for a rectangle : A xy= At this point we can either leave the equation as is and differentiate it or we can plug in 3y x= to simplify the equation down to a single variable then differentiate. Doing this gives,


Calculus I


( ) 23A x x= Step 2 Now we need to differentiate with respect to t. If we use the equation in terms of only x, which is probably the easiest to use we get, 6A x x′ ′=

If we use the equation in terms of both x and y we get, A x y x y′ ′ ′= + Step 3 Now all we need to do is plug in the known quantities and solve for A′ . Using the equation in terms of only x is the “easiest” because we already have all the known quantities from the problem statement itself. Doing this gives,

( )( )6 6 2 72A′ = − = − Now let’s use the equation in terms of x and y. We know that 6x = and 2x′ = − from the

problem statement. From part (a) we have 6y′ = − and we also know that ( )3 6 18y = = .

Using these gives,

( ) ( ) ( )( )6 6 2 18 72A′ = − + − = − So, as we can see both gives the same result but the second method is slightly more work, although not much more. 4. A thin sheet of ice is in the form of a circle. If the ice is melting in such a way that the area of the sheet is decreasing at a rate of 0.5 m2/sec at what rate is the radius decreasing when the area of the sheet is 12 m2 ? Step 1 We’ll call the area of the sheet A and the radius r and we know that the area of a circle is given by,

2A rπ= We know that 0.5A′ = − and want to determine r′ when 12A = . Step 2 Next step is to simply differentiate the equation with respect to t. 2A r rπ′ ′=


Calculus I


Step 3 Now, to finish this problem off we’ll first need to go back to the equation of the area and use the fact that we know the area at the point we are interested in and determine the radius at that time.

2 1212 1.9544r rππ

= ⇒ = =

The rate of change of the radius is then,

( )0.5 2 1.9544 0.040717r rπ ′ ′− = ⇒ = − 5. A person is standing 350 feet away from a model rocket that is fired straight up into the air at a rate of 15 ft/sec. At what rate is the distance between the person and the rocket increasing (a) 20 seconds after liftoff? (b) 1 minute after liftoff? Step 1 Here is a sketch for this situation that will work for both parts so we’ll put it here.

Step 2 In both parts we know that 15y′ = and want to determine z′ for each given time. Using the Pythagorean Theorem we get the following equation to relate y and z. 2 2 2 2350 122500z y y= + = + Step 3 Finally, let’s differentiate this with respect to t and we can even solve it for z′ so the actual solution will be quick and simple to find.

2 2 y yz z y y zz

′′ ′ ′= ⇒ =

We have now reached a point where the process will differ for each part.


Calculus I


(a) At what rate is the distance between the person and the rocket increasing 20 seconds after liftoff? To finish off this problem all we need to do is determine y (from the speed of the rocket and given time) and z (reusing the Pythagorean Theorem).

( )( ) 2 215 20 300 300 350 212500 50 85 460.9772y z= = = + = = =

The rate of change of the distance between the two is then,

( )( )300 159.76187

460.9772z′ = =

(b) At what rate is the distance between the person and the rocket increasing 1 minute after liftoff? This part is nearly identical to the first part with the exception that the time is now 60 seconds (and note that we MUST be in seconds because the speeds are in time of seconds). Here is the work for this problem.

( ) ( )

( ) ( )

2 215 60 900 900 350 932500 50 373 965.6604

900 1513.98007

965.6604

y z

z

= = = + = = =

′ = =

6. A plane is 750 meters in the air flying parallel to the ground at a speed of 100 m/s and is initially 2.5 kilometers away from a radar station. At what rate is the distance between the plane and the radar station changing (a) initially and (b) 30 seconds after it passes over the radar station? See the (probably bad) sketch below to help visualize the problem.

(a) At what rate is the distance between the plane and the radar station changing initially? Step 1


Calculus I


For this part we know that 100x′ = − when 2500x = . In this case note that x′ must be negative because x will be decreasing in this part. Also note that we converted x to meters since all the other quantities are in meters. Here is a sketch for this part.

Step 2 We want to determine z′ in this part so using the Pythagorean Theorem we get the following equation to relate x and z. 2 2 2 2750 562500z x x= + = + Step 3 Finally, let’s differentiate this with respect to t and we can even solve it for z′ so the actual solution will be quick and simple to find.

2 2 x xz z x x zz

′′ ′ ′= ⇒ =

Step 4 To finish off this problem all we need to do is determine z (reusing the Pythagorean Theorem) and then plug into the equation from Step 3 above.

2 22500 750 6812500 250 109 2610.0766z = + = = = The rate of change of the distance between the two for this part is,

( )( )2500 10095.7826

2610.0766z

−′ = = −

(b) At what rate is the distance between the plane and the radar station changing 30 seconds after it passes over the radar station? Step 1 For this part we know that 100x′ = and it will be positive in this case because x will now be increasing as we can see in the sketch below.


Calculus I


Step 2 As with the previous part we want to determine z′ and equation we’ll need is identical to the previous part so we’ll just rewrite both it and its derivative here.

2 2 2 2750 562500

2 2

z x xx xz z x x zz

= + = +′

′ ′ ′= ⇒ =

Step 3 To finish off this problem all we need to do is determine both x and z. For x we know the speed of the plane and the fact that it has flown for 30 seconds after passing over the radar station. So x is, ( )( )100 30 3000x = = For z we just need to reuse the Pythagorean Theorem.

2 23000 750 9562500 750 17 3092.3292z = + = = = The rate of change of the distance between the two for this part is then,

( ) ( )3000 10097.0143

3092.3292z′ = =

7. Two people are at an elevator. At the same time one person starts to walk away from the elevator at a rate of 2 ft/sec and the other person starts going up in the elevator at a rate of 7 ft/sec. What rate is the distance between the two people changing 15 seconds later? Step 1 Here is a sketch for this part.


Calculus I


We want to determine z′ after 15 seconds given that 2x′ = , 7y′ = and assuming that they start at the same point. Step 2 Hopefully it’s clear that we’ll need the Pythagorean Theorem to solve this problem so here is that. 2 2 2z x y= + Step 3 Finally, let’s differentiate this with respect to t and we can even solve it for z′ so the actual solution will be quick and simple to find.

2 2 2 x x y yz z x x y y zz

′ ′+′ ′ ′ ′= + ⇒ =

Step 4 To finish off this problem all we need to do is determine all three lengths of the triangle in the sketch above. We can find x and y using their speeds and time while we can find z by reusing the Pythagorean Theorem.

( )( ) ( )( )2 2

2 15 30 7 15 105

30 105 11925 15 53 109.2016

x y

z

= = = =

= + = = =

The rate of change of the distance between the two people is then,

( )( ) ( )( )30 2 105 77.2801

109.2016z

+′ = =

8. Two people on bikes are at the same place. One of the bikers starts riding directly north at a rate of 8 m/sec. Five seconds after the first biker started riding north the second starts to ride directly east at a rate of 5 m/sec. At what rate is the distance between the two riders increasing 20 seconds after the second person started riding?


Calculus I


Step 1 Here is a sketch of this situation.

We want to determine z′ after 20 seconds after the second biker starts riding east given that

5x′ = , 8y′ = and assuming that they start at the same point. Step 2 Hopefully it’s clear that we’ll need the Pythagorean Theorem to solve this problem so here is that. 2 2 2z x y= + Step 3 Finally, let’s differentiate this with respect to t and we can even solve it for z′ so the actual solution will be quick and simple to find.

2 2 2 x x y yz z x x y y zz

′ ′+′ ′ ′ ′= + ⇒ =

Step 4 To finish off this problem all we need to do is determine all three lengths of the triangle in the sketch above. We can find x and y using their speeds and time while we can find z by reusing the Pythagorean Theorem. Note that the biker riding east will be riding for 20 seconds and the biker riding north will be riding for 25 seconds (this biker started 5 seconds earlier…).

( ) ( ) ( )( )2 2

5 20 100 8 25 200

100 200 50000 100 5 223.6068

x y

z

= = = =

= + = = =

The rate of change of the distance between the two people is then,

( ) ( ) ( )( )100 5 200 89.3915

223.6068z

+′ = =

9. A light is mounted on a wall 5 meters above the ground. A 2 meter tall person is initially 10 meters from the wall and is moving towards the wall at a rate of 0.5 m/sec. After 4 seconds of moving is the tip of the shadow moving (a) towards or away from the person and (b) towards or away from the wall?


Calculus I


Step 1 Here is a sketch for this situation that will work for both parts so we’ll put it here. Also note that we know that 0.5px′ = − for both parts.

(a) After 4 seconds of moving is the tip of the shadow towards or away from the person? Step 2 In this case we want to determine sx′ when ( )10 4 0.5 8px = − = (although it will turn out that

we simply don’t need this piece of information for this problem….). We can use the idea of similar triangles to get the following equation.

25

s s

p s

x xx x x

= =+

If we solve this for sx we arrive at,

( )2

5

2 2 25 5 3 p

p s s

p s s s

x x x

x x x x x

+ =

+ = ⇒ =

This equation will work perfectly for us. Step 3 Differentiation with respect to t will give us, 2

3 psx x′ ′= Step 4 Finishing off this problem is very simple as all we need to do is plug in the known speed.

( )2 13 30.5sx′ = − = −


Calculus I


Because this rate is negative we can see that the tip of the shadow is moving towards the person at a rate of 1

3 m/s. (b) After 4 seconds of moving is the tip of the shadow towards or away from the wall? Step 2 In this case we want to determine x′ and the equation is really simple. All we need is, p sx x x= + Step 3 Differentiation with respect to t will give us, p sx x x′ ′ ′= + Step 4 Finishing off this problem is very simple as all we need to do is plug in the known speeds and note that we will need to result from the first part here. So we have 1

2px′ = − from the problem

statement and 13sx′ = − from the previous part.

( ) 51 12 3 6x′ = − + − = −

Because this rate is negative we can see that the tip of the shadow is moving towards the wall at a rate of 5

6 m/s. 10. A tank of water in the shape of a cone is being filled with water at a rate of 12 m3/sec. The base radius of the tank is 26 meters and the height of the tank is 8 meters. At what rate is the depth of the water in the tank changing with the radius of the top of the water is 10 meters? Step 1 Here is a sketch of the cross section of the tank and it is not even remotely to scale as I found it easier to reuse an old image that I had lying around. I can be a little lazy sometimes….


Calculus I


We want to determine h′ when 10r = and we know that 12V ′ = . Step 2 We’ll need the equation for the volume of a cone. 21

3V r hπ= This is a problem however as it has both r and h in it and it would be best to have only h since we need h′ . We can use similar triangles to fix this up. Based on similar triangles we get the following equation which can be solved for r.

134

268

r r hh

= ⇒ =

Plugging this into the volume equation gives, 3169

48V hπ= Step 3 Next, let’s differentiate this with respect to t. 2169

16V h hπ′ ′= Step 4 To finish off this problem all we need to do is determine the value of h for the time we are interested in. This can easily be done from the similar triangle equation and the fact that we know 10r = .

( ) 404 413 13 1310h r= = =

The rate of change of the height of the water is then,


Calculus I


( )2169 40 316 13 2512 100h h h ππ π′ ′ ′= = ⇒ =

11. The angle of elevation is the angle formed by a horizontal line and a line joining the observer’s eye to an object above the horizontal line. A person is 500 feet way from the launch point of a hot air balloon. The hot air balloon is starting to come back down at a rate of 15 ft/sec. At what rate is the angle of elevation, θ , changing when the hot air balloon is 200 feet above the ground. See the (probably bad) sketch below to help visualize the angle of elevation if you are having trouble seeing it.

Step 1 Putting variables and known quantities on the sketch from the problem statement gives,

We want to determine θ ′ when 200y = and we know that 15y′ = − . Step 2 There are a variety of equations that we could use here but probably the best one that involves all of the known and needed quantities is,

( )tan500

yθ =

Step 3 Differentiating with respect to t gives,

( ) ( )2 2sec cos500 500y y

θ θ θ θ′ ′

′ ′= ⇒ =

Step 4


Calculus I


To finish off this problem all we need to do is determine the value of θ for the time in question. We can either use the original equation to do this or we could acknowledge that all we really need is ( )cos θ and we could do a little right triangle trig to determine that.

For this problem we’ll just use the original equation to find the value of θ .

( ) ( )1200 2500 5tan tan 0.38051radiansθ θ −= ⇒ = =

The rate of change of the angle of elevation is then,

( )215 cos 0.38051 0.2586500

θ−′ = = −

Higher Order Derivatives 1. Determine the fourth derivative of ( ) 7 4 33 6 8 12 18h t t t t t= − + − +

Step 1 Not much to this problem other than to take four derivatives so each step will show each successive derivatives until we get to the fourth. The first derivative is then, ( ) 6 3 221 24 24 12h t t t t′ = − + − Step 2 The second derivative is, ( ) 5 2126 72 48h t t t t′′ = − +

Step 3 The third derivative is, ( ) 4630 144 48h t t t′′′ = − + Step 4 The fourth, and final derivative for this problem, is,

( ) ( )4 32520 144h t t= −


Calculus I


2. Determine the fourth derivative of ( ) 3 2 1V x x x x= − + −

Step 1 Not much to this problem other than to take four derivatives so each step will show each successive derivatives until we get to the fourth. The first derivative is then, ( ) 23 2 1V x x x′ = − + Step 2 The second derivative is, ( ) 6 2V x x′′ = −

Step 3 The third derivative is, ( ) 6V x′′′ = Step 4 The fourth, and final derivative for this problem, is,

( ) ( )4 0V x = Note that we could have just as easily used the Fact from the notes to arrive at this answer in one step.

3. Determine the fourth derivative of ( ) 5 32

148

f x x xx

= − −

Step 1 Not much to this problem other than to take four derivatives so each step will show each successive derivatives until we get to the fourth. After a quick rewrite of the function to help with the differentiation the first derivative is,

( ) ( )3 21 1

2 35 52 21 12 1 148 5 4 2

f x x x x f x x x x− −− −′= − − ⇒ = + −

Step 2 The second derivative is,

( )7 3

45 224 3 125 4 4

f x x x x− −−′′ = − − +

Step 3 The third derivative is,


Calculus I


( )12 5

55 2168 33125 8

f x x x x− −−′′′ = + −

Step 4 The fourth, and final derivative for this problem, is,

( ) ( )17 7

4 65 22016 1515625 16

f x x x x− −−= − − +

4. Determine the fourth derivative of ( ) ( ) ( )37sin cos 1 2wf w w= + −

Step 1 Not much to this problem other than to take four derivatives so each step will show each successive derivatives until we get to the fourth. The first derivative is then,

( ) ( ) ( )37 cos 2sin 1 23

wf w w′ = + −


( ) ( ) ( )37 sin 4 cos 1 29

wf w w′′ = − − −


( ) ( ) ( )37 cos 8sin 1 227

wf w w′′′ = − − −


( ) ( ) ( ) ( )43

7 sin 16cos 1 281

wf w w= + −

5. Determine the fourth derivative of ( )5 48ln 2zy z−= +e

Step 1 Not much to this problem other than to take four derivatives so each step will show each successive derivatives until we get to the fourth. The first derivative is then,


Calculus I


3

5 5 14

85 8 5 322

z zdy z zdx z

− − − = − + = − +

e e


2

5 22 25 32zd y z

dx− −= −e


3

5 33 125 64zd y z

dx− −= − +e


4

5 44 625 192zd y z

dx− −= −e

6. Determine the second derivative of ( ) ( )3sin 2 9g x x x= −

Step 1 Not much to this problem other than to take two derivatives so each step will show each successive derivatives until we get to the second. The first derivative is then,

( ) ( ) ( )2 36 9 cos 2 9g x x x x′ = − − Step 2 Do not forget that often we will end up needing to do a product rule in the second derivative even though we did not need to do that in the first derivative. The second derivative is then,

( ) ( ) ( ) ( )23 2 312 cos 2 9 6 9 sin 2 9g x x x x x x x′′ = − − − −

7. Determine the second derivative of ( )3ln 7z x= −

Step 1 Not much to this problem other than to take two derivatives so each step will show each successive derivatives until we get to the second. The first derivative is then,


Calculus I


2

3

37

dz xdx x

−=

−

Step 2 Do not forget that often we will end up needing to do a quotient rule in the second derivative even though we did not need to do that in the first derivative. The second derivative is then,

( ) ( )( )

( ) ( )

3 2 22 4

2 22 3 3

6 7 3 3 42 3

7 7

x x x xd z x xdx x x

− − − − − − −= =

− −

8. Determine the second derivative of ( )( )42

2

6 2Q v

v v=

+ −

Step 1 Not much to this problem other than to take two derivatives so each step will show each successive derivatives until we get to the second. We’ll do a quick rewrite of the function to help with the derivatives and then the first derivative is,

( ) ( )( ) ( )( )

42

52

2 6 2

8 2 2 6 2

Q v v v

Q v v v v

−

−

= + −

′ = − − + −

Step 2 Do not forget that often we will end up needing to do a product rule in the second derivative even though we did not need to do that in the first derivative. The second derivative is then,

( ) ( ) ( ) ( )5 622 216 6 2 40 2 2 6 2Q v v v v v v− −

′′ = + − + − + − 9. Determine the second derivative of ( ) ( )2cos 7H t t=

Step 1 Not much to this problem other than to take two derivatives so each step will show each successive derivatives until we get to the second. The first derivative is then, ( ) ( ) ( )14cos 7 sin 7H t t t′ = − Step 2 Do not forget that often we will end up needing to do a product rule in the second derivative even though we did not need to do that in the first derivative. The second derivative is then,


Calculus I


( ) ( ) ( ) ( ) ( ) ( ) ( )2 298sin 7 sin 7 98cos 7 cos 7 98sin 7 98cos 7H t t t t t t t′′ = − = − Note that, in this case, if we recall our trig formulas we could have reduced the product in the first derivative to a single trig function which would have then allowed us to avoid the product rule for the second derivative. Can you figure out what the formula is? 10. Determine the second derivative of 3 22 1 4x y y+ = − Step 1 Not much to this problem other than to take two derivatives so each step will show each successive derivatives until we get to the second. Note however that we are going to have to do implicit differentiation to do each derivative. Here is the work for the first derivative. If you need a refresher on implicit differentiation go back to that section and check some of the problems in that section.

( )

2

2 22

6 2 46 32 4 6

2 4 2

x y y yx xy y x y

y y

′ ′+ = −

− −′ ′+ = − ⇒ = =+ +

Step 2 Now, the second derivative will also need implicit differentiation. Note as well that we can work with the first derivative in its present form which will require the quotient rule or we can rewrite it as,

( ) 123 2y x y −′ = − + and use the product rule. These get messy enough as it is so we’ll go with the product rule to try and keep the “mess” down a little. Using implicit differentiation to take the derivative of first derivative gives,

( ) ( ) ( )1 226 2 3 2dy y x y x y ydx

− −′′ ′ ′= = − + + + Step 3 Finally, recall that we don’t want a y′ in the second derivative so to finish this out we need to plug in the formula for y′ (which we know…) and do a little simplifying to get the final answer.

( ) ( ) ( )( ) ( ) ( )1 2 1 1 32 2 46 2 3 2 3 2 6 2 9 2y x y x y x y x y x y− − − − −′′ = − + + + − + = − + − +


Calculus I


11. Determine the second derivative of 26 1y xy− = Step 1 Not much to this problem other than to take two derivatives so each step will show each successive derivatives until we get to the second. Note however that we are going to have to do implicit differentiation to do each derivative. Here is the work for the first derivative. If you need a refresher on implicit differentiation go back to that section and check some of the problems in that section.

( )

2

22

6 2 0

6 26 2

y y xy yyxy y y y

xy

′ ′− − =

′ ′− = ⇒ =−

Step 2 Now, the second derivative will also need implicit differentiation. Note as well that we can work with the first derivative in its present form which will require the quotient rule or we can rewrite it as,

( ) 12 6 2y y xy −′ = − and use the product rule. These get messy enough as it is so we’ll go with the product rule to try and keep the “mess” down a little. Using implicit differentiation to take the derivative of first derivative gives,

( ) ( ) ( ) ( )1 222 6 2 6 2 2 2dy y y y xy y xy y xydx

− −′′ ′ ′ ′= = − − − − − Step 3 Finally, recall that we don’t want a y′ in the second derivative. So, to finish this out let’s do a little “simplifying” of the to make it “easier” to plug in the formula for y′ .

( ) ( ) ( )( ) ( )( ) ( )

1 2 23 2

1 1 23

2 6 2 2 6 2 2 6 2

2 6 2 1 6 2 2 6 2

y y y xy y xy xy y xy

y y xy xy xy y xy

− − −

− − −

′′ ′ ′= − + − + −

′= − + − + −

The point of all of this was to get down to a single y′ in the formula for the second derivative, which won’t always be possible to do, and a little factoring to try and make things a little easier to deal with. Finally all we need to do is plug in the formula for y′ to get the final answer.


Calculus I


( ) ( ) ( )( ) ( )

( ) ( )( ) ( )

1 1 1 22 3

2 1 23 3

2 6 2 6 2 1 6 2 2 6 2

2 6 2 1 6 2 2 6 2

y y y xy xy xy xy y xy

y xy xy xy y xy

− − − −

− − −

′′ = − − + − + −

= − + − + −

Note that for a further simplification step, if we wanted to go further, we could factor a

( ) 232 6 2y xy −− out of both terms to get,

( ) ( )( )2 132 6 2 2 6 2y y xy xy xy− −′′ = − + −

Logarithmic Differentiation 1. Use logarithmic differentiation to find the first derivative of

( ) ( )72 25 3 6 8 12f x x x x= − + − .

Step 1 Take the logarithm of both sides and do a little simplifying.

( ) ( )

( ) ( )( ) ( )

12

72 2

72 2

2 212

ln ln 5 3 6 8 12

ln 5 3 ln 6 8 12

7ln 5 3 ln 6 8 12

f x x x x

x x x

x x x

= − + − = − + + −

= − + + −

Step 2 Use implicit differentiation to differentiate both sides with respect to x.

( )( ) 2 2 2 2

6 1 12 8 42 6 475 3 2 6 8 12 5 3 6 8 12

f x x x x xf x x x x x x x′ − + − +

= + = +− + − − + −

Step 3 Finally, solve for the derivative and plug in the equation for ( )f x .


Calculus I


( ) ( )

( )

2 2

72 22 2

42 6 45 3 6 8 12

42 6 45 3 6 8 125 3 6 8 12

x xf x f xx x x

x xx x xx x x

− + ′ = + − + −

− + = − + − + − + −

2. Use logarithmic differentiation to find the first derivative of ( )

( )

2

34

sin 3

6

z zy

z

+=

−.


( ) ( )

( )( ) ( )

( )

232 4

34

2 4

sin 3ln ln ln sin 3 ln 6

6

ln sin 3 3ln 6

z zy z z z

z

z z z

+ = = + − − − = + − −

Step 2 Use implicit differentiation to differentiate both sides with respect to z.

( ) ( )

( ) ( ) ( )2 3 3

24 42

3 2 cos 3 4 123 3 2 cot 36 6sin 3

z z zy z zz z zy z zz z

+ +′ −= − = + + + − −+

Step 3 Finally, solve for the derivative and plug in the equation for y .

( ) ( ) ( )( )

( ) ( )23 3

2 234 44

sin 312 123 2 cot 3 3 2 cot 36 66

z zz zy y z z z z z zz zz

+ ′ = + + + = + + + − − −

3. Use logarithmic differentiation to find the first derivative of ( ) ( )3

24

5 8 1 9cos 4

10

t th t

t t

+ −=

+.



Calculus I


( ) ( )

( )

( ) ( )( ) ( )

( ) ( )( ) ( )

3

24

243

1112 432

21 1 12 3 4

5 8 1 9cos 4ln ln

10

ln 5 8 1 9cos 4 ln 10

ln 5 8 ln 1 9cos 4 ln 10

ln 5 8 ln 1 9cos 4 10

t th t

t t

t t t t

t t t t

t t t t

+ − = +

= + − − + = + + − − +

= + + − − +

Note that the logarithm simplification work was a little complicated for this problem, but if you know your logarithm properties you should be okay with that. Step 2 Use implicit differentiation to differentiate both sides with respect to t.

( )( )

( )( )

1 1 12 3 4 2

36sin 45 2 105 8 1 9cos 4 10

h t t th t t t t t′ +

= + −+ − +

Step 3 Finally, solve for the derivative and plug in the equation for ( )h t .

( ) ( ) ( )( )

( ) ( )( )

5 512 2 2

2

3 5 512 2 2

224

12sin 45 8 1 9cos 4 10

5 8 1 9cos 4 12sin 45 8 1 9cos 4 1010

t th t h tt t t t

t t t tt t t tt t

+′ = + − + − +

+ − += + − + − ++

4. Find the first derivative of ( ) ( )43 7 wg w w= − .

Step 1 We just need to do some logarithmic differentiation so take the logarithm of both sides and do a little simplifying.

( ) ( ) ( )4ln ln 3 7 4 ln 3 7wg w w w w = − = −

Step 2 Use implicit differentiation to differentiate both sides with respect to w. Don’t forget to product rule the right side.

( )( ) ( ) ( )3 124ln 3 7 4 4ln 3 7

3 7 3 7g w ww w wg w w w′

= − + = − +− −


Calculus I


Step 3 Finally, solve for the derivative and plug in the equation for ( )g w .

( ) ( ) ( )

( ) ( )4

124ln 3 73 7

123 7 4ln 3 73 7

w

wg w g w ww

ww ww

′ = − + −

= − − + −

5. Find the first derivative of ( ) ( ) ( )sin 282xxf x x= − e .

Step 1 We just need to do some logarithmic differentiation so take the logarithm of both sides and do a little simplifying.

( ) ( ) ( ) ( ) ( )sin 28 8ln ln 2 sin 2 ln 2xx xf x x x x = − = −

e e

Step 2 Use implicit differentiation to differentiate both sides with respect to x. Don’t forget to product rule the right side.

( )( ) ( ) ( ) ( )

( ) ( ) ( )

88

8

88

8

2 82cos 2 ln 2 sin 22

2 82cos 2 ln 2 sin 22

xx

x

xx

x

f xx x x

f x x

x x xx

′ −= − +

−

−= − +

−

eee

eee

Step 3 Finally, solve for the derivative and plug in the equation for ( )f x .

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

88

8

8sin 28 88

2 82cos 2 ln 2 sin 22

2 82 2cos 2 ln 2 sin 22

x

xx

x

xx x

x

f x f x x x xx

x x x xx

−′ = − + −

−= − − + −

eee

ee ee


CALCULUS I Assignment Problems

Derivatives

Paul Dawkins

Calculus I



Introduction .............................................................................................................................................. 1 The Definition of the Derivative .............................................................................................................. 2 Interpretations of the Derivative ............................................................................................................. 4 Differentiation Formulas .......................................................................................................................... 8 Product and Quotient Rule ..................................................................................................................... 11 Derivatives of Trig Functions ................................................................................................................ 12 Derivatives of Exponential and Logarithm Functions ......................................................................... 15 Derivatives of Inverse Trig Functions ................................................................................................... 16 Derivatives of Hyperbolic Functions ..................................................................................................... 17 Chain Rule ............................................................................................................................................... 17 Implicit Differentiation........................................................................................................................... 21 Related Rates .......................................................................................................................................... 24 Higher Order Derivatives ....................................................................................................................... 27 Logarithmic Differentiation ................................................................................................................... 29

Preface Here are a set of problems for my Calculus I notes. These problems do not have any solutions available on this site. These are intended mostly for instructors who might want a set of problems to assign for turning in. I try to put up both practice problems (with solutions available) and these problems at the same time so that both will be available to anyone who wishes to use them. As with the set of practice problems I write these as I get the time and some sections will have only a few problems at this point and others won’t have any problems in them yet. Rest assured that I’m always trying to get more problems written but this site has been written and maintained in my spare time and so I usually cannot devote as much time as I’d like to the site.

Derivatives

Introduction Here are a set of problems for which no solutions are available. The main intent of these problems is to have a set of problems available for any instructors who are looking for some extra problems.


Calculus I


Note that some sections will have more problems than others and some will have more or less of a variety of problems. Most sections should have a range of difficulty levels in the problems although this will vary from section to section. Here is a list of topics in this chapter that have problems written for them. The Definition of the Derivative Interpretation of the Derivative Differentiation Formulas Product and Quotient Rule Derivatives of Trig Functions Derivatives of Exponential and Logarithm Functions Derivatives of Inverse Trig Functions Derivatives of Hyperbolic Functions Chain Rule Implicit Differentiation Related Rates Higher Order Derivatives Logarithmic Differentiation

The Definition of the Derivative Use the definition of the derivative to find the derivative of the following functions. 1. ( ) 10g x =

2. ( ) 8T y = −

3. ( ) 5 7f x x= +

4. ( ) 1 12Q t t= −

5. ( ) 2 3f z z= +

6. ( ) 2 8 20R w w w= − +

7. ( ) 26V t t t= −

8. ( ) 22 8 10Q t t t= − +


Calculus I


9. ( ) 21 10 7g z z z= + −

10. ( ) 35f x x x= −

11. ( ) 32 9 5Y t t t= + +

12. ( ) 3 22Z x x x x= − −

13. ( ) 23

f tt

=−

14. ( ) 21xg x

x+

=−

15. ( )2

2tQ t

t=

+

16. ( ) 8f w w= +

17. ( ) 14 3V t t= +

18. ( ) 2 5G x x= −

19. ( ) 1 4Q t t= +

20. ( ) 2 1f x x= +

21. ( ) 1W tt

=

22. ( ) 41

g xx

=−

23. ( )f x x x= +


Calculus I


24. ( ) 1f x xx

= +

Interpretations of the Derivative

For problems 1 – 3 use the graph of the function, ( )f x , estimate the value of ( )f a′ for the

given values of a. 1. (a) 5a = − (b) 1a =

2. (a) 2a = − (b) 3a =

3. (a) 3a = − (b) 4a =


Calculus I


For problems 4 – 6 sketch the graph of a function that satisfies the given conditions. 4. ( )7 5f − = , ( )7 3f ′ − = − , ( )4 1f = − , ( )4 1f ′ =

5. ( )1 2f = , ( )1 4f ′ = , ( )6 2f = , ( )6 3f ′ =

6. ( )1 9f − = − , ( )1 0f ′ − = , ( )2 1f = − , ( )2 3f ′ = , ( )5 4f = , ( )5 1f ′ = −

For problems 7 – 9 the graph of a function, ( )f x , is given. Use this to sketch the graph of the

derivative, ( )f x′ .

7.

8.


Calculus I


9.

10. Answer the following questions about the function ( ) 21 10 7g z z z= + − .

(a) Is the function increasing or decreasing at 0z = ? (b) Is the function increasing or decreasing at 2z = ? (c) Does the function ever stop changing? If yes, at what value(s) of z does the function stop changing? 11. What is the equation of the tangent line to ( ) 35f x x x= − at 1x = .

12. The position of an object at any time t is given by ( ) 22 8 10s t t t= − + .

(a) Determine the velocity of the object at any time t. (b) Is the object moving to the right or left at 1t = ? (c) Is the object moving to the right or left at 4t = ? (d) Does the object ever stop moving? If so, at what time(s) does the object stop moving?


Calculus I


13. Does the function ( ) 2 8 20R w w w= − + ever stop changing? If yes, at what value(s)

of w does the function stop changing? 14. Suppose that the volume of air in a balloon for 0 6t≤ ≤ is given by ( ) 26V t t t= − .

(a) Is the volume of air increasing or decreasing at 2t = ? (b) Is the volume of air increasing or decreasing at 5t = ? (c) Does the volume of air ever stop changing? If yes, at what times(s) does the volume stop changing? 15. What is the equation of the tangent line to ( ) 5 7f x x= + at 4x = − ?

16. Answer the following questions about the function ( ) 3 22Z x x x x= − − .

(a) Is the function increasing or decreasing at 1x = − ? (b) Is the function increasing or decreasing at 2x = ? (c) Does the function ever stop changing? If yes, at what value(s) of x does the function stop changing?

17. Determine if the function ( ) 14 3V t t= + increasing or decreasing at the given points.

(a) 0t = (b) 5t = (c) 100t =

18. Suppose that the volume of water in a tank for 0t ≥ is given by ( )2

2tQ t

t=

+ .

(a) Is the volume of water increasing or decreasing at 0t = ? (b) Is the volume of water increasing or decreasing at 3t = ? (c) Does the volume of water ever stop changing? If so, at what times(s) does the volume stop changing? 19. What is the equation of the tangent line to ( ) 10g x = at 16x = ?

20. The position of an object at any time t is given by ( ) 1 4Q t t= + .

(a) Determine the velocity of the object at any time t. (b) Does the object ever stop moving? If so, at what time(s) does the object stop moving? 21. Does the function ( ) 32 9 5Y t t t= + + ever stop changing? If yes, at what value(s)

of t does the function stop changing?


Calculus I


Differentiation Formulas For problems 1 – 20 find the derivative of the given function. 1. ( ) 3 88 4 2g x x x= − +

2. ( ) 10 5 3 27 2f z z z z z= − + −

3. 4 38 10 9 4y x x x= − − + 4. ( ) 4 43 3f x x x x−= + −

5. ( ) 10 109 8 12R t t t−= + +

6. ( ) 6 3 13 8 9h y y y y− − −= − +

7. ( ) 7 3 2 42 6 8 1g t t t t t− − −= + − + −

8. 6 47 3z x x x= − +

9. ( ) 9 34 7 427 2f x x x x= − +

10. ( ) 5629

76h y y yy

= + +

11. ( ) 2 5

4 1 17 2

g zz z z

= + −

12. 2 39 3

2 1 93 7

y t tt t

= + − −

13. ( ) 36 5 2

1 1W x xx x

= − +

14. ( ) ( )( )25 1g w w w= − +


Calculus I


15. ( ) ( )31 9h x x x= −

16. ( ) ( )233 2f t t= −

17. ( ) ( )( )21 2 2g t x x x= + − +

18. 24 8 2x xy

x− +

=

19. ( )4 2

3

2 7t t tY tt

− +=

20. ( ) ( )2 523

w w wS w

w− +

=

For problems 21 – 26 determine where, if anywhere, the function is not changing. 21. ( ) 3 22 9 108 14f x x x x= − − +

22. ( ) 2 3 445 300 20 3u t t t t= + + −

23. ( ) 3 29 10Q t t t t= − + −

24. ( ) 3 22 3 4 5h w w w w= + + +

25. ( ) 2 3 49 8 3g x x x x= + + −

26. ( ) ( )22 1G z z z= −

27. Find the tangent line to ( ) 5 23 4 9 12f x x x x= − + − at 1x = − .

28. Find the tangent line to ( )2 1xg xx+

= at 2x = .

29. Find the tangent line to ( ) 42 8h x x x= − at 16x = .


Calculus I


30. The position of an object at any time t is given by ( ) 4 3 23 44 108 20s t t t t= − + + .

(a) Determine the velocity of the object at any time t. (b) Does the object ever stop changing? (c) When is the object moving to the right and when is the object moving to the left? 31. The position of an object at any time t is given by ( ) 3 4 51 150 45 2s t t t t= − + − .

(a) Determine the velocity of the object at any time t. (b) Does the object ever stop changing? (c) When is the object moving to the right and when is the object moving to the left? 32. Determine where the function ( ) 3 24 18 336 27f x x x x= − − + is increasing and decreasing.

33. Determine where the function ( ) 4 3 22 15 9g w w w w= + − − is increasing and decreasing.

34. Determine where the function ( ) 3 224 192 50V t t t t= − + − is increasing and decreasing.

35. Determine the percentage of the interval [ ]6,4− on which ( ) 3 4 57 10 5 2f x x x x= + − − is

increasing. 36. Determine the percentage of the interval [ ]5,2− on which ( ) 4 3 23 8 144f x x x x= − − is

decreasing. 37. Is ( ) 2 33 2h z x x x= − + + increasing or decreasing more on the interval [ ]1,1− ?

38. Determine where, if anywhere, the tangent line to ( ) 212 9 3f x x x= − + is parallel to the

line 1 7y x= − . 39. Determine where, if anywhere, the tangent line to ( ) 2 38 4 2f x x x x= + + − is perpendicular

to the line 1 84 3

y x= − + .

40. Determine where, if anywhere, the tangent line to ( ) 3 8f x x x= − is perpendicular to the

line 2 11y x= − .

41. Determine where, if anywhere, the tangent line to ( ) 13 19xf x

x= + is parallel to the line

y x= .


Calculus I


Product and Quotient Rule For problems 1 – 7 use the Product Rule or the Quotient Rule to find the derivative of the given function.

1. ( ) ( )( )3 22 3 8h z z z= − +

2. ( ) ( )32 7 2f x x xx

= − −

3. ( )( )2 35 1 12 2y x x x x= − + + −

4. ( )3

21xg xx

=+

5. ( )24

6y yZ y

y−

=−

6. ( )2

3

1 105 2

t tV tt t

− +=

+

7. ( ) ( )( )1 4 23 9w w

f ww

− +=

+

For problems 8 – 12 use the fact that ( )3 12f − = , ( )3 9f ′ − = , ( )3 4g − = − , ( )3 7g′ − = ,

( )3 2h − = − and ( )3 5h′ − = determine the value of the indicated derivative.

8. ( ) ( )3f g ′ −

9. ( )3hg

′ −


Calculus I


10. ( )3f gh

′ −

11. If ( ) ( )y x f x h x= − determine 3x

dydx =−

.

12. If ( ) ( )

( )1 g x h x

yx f x

−=

+ determine

3x

dydx =−

.

13. Find the equation of the tangent line to ( ) ( )( )2 28 1f x x x x= − + + at 2x = − .

14. Find the equation of the tangent line to ( )3

2

42xf x

x x−

=+

at 1x = .

15. Determine where ( ) 2

212

zg zz

−=

+ is increasing and decreasing.

16. Determine where ( ) ( )( )23 1 2R x x x x= − − + is increasing and decreasing.


2

71 2

t th tt

−=

+ is increasing and decreasing.

18. Determine where ( ) 11

xf xx

+=

− is increasing and decreasing.

19. Using the Product Rule for two functions prove the Product Rule for three functions.

( )f g h f g h f g h f g h′ ′ ′ ′= + +

Derivatives of Trig Functions For problems 1 – 6 evaluate the given limit.

1. ( )0

3limsint

tt→

2. ( )

0

sin 9lim

10w

ww→


Calculus I


3. ( )( )0

sin 2lim

sin 17θ

θθ→

4. ( )

0

sin 4lim

3 12x

xx→

++

5. ( )

0

cos 1lim

9x

xx→

−

6. ( )

0

cos 8 1lim

2z

zz→

−

For problems 6 – 10 differentiate the given function. 6. ( ) ( ) ( )4 9sin 2 tanh x x x x= − +

7. ( ) ( ) ( ) ( )8sec cos 4cscg t t t t= + −

8. ( ) ( )6cot 8cos 9y w w= − +

9. ( ) ( ) ( )8sec cscf x x x=

10. ( ) ( )98 tanh t t t= −

11. ( ) ( )5 26 8 sinR x x x x= +

12. ( ) ( )3

cos3

zh z z

z= −

13. ( ) ( )( )

1 cos1 sin

xY x

x+

=−

14. ( ) ( )( )

sec3

1 9 tanw

f w ww

= −+


Calculus I


15. ( ) ( )2

cot1

t tg t

t=

+

16. Find the tangent line to ( ) ( )2 tan 4f x x x= − at 0x = .

17. Find the tangent line to ( ) ( )secf x x x= at 2x π= .

18. Find the tangent line to ( ) ( ) ( )cos secf x x x= + at x π= .

19. The position of an object is given by ( ) ( ) ( )9sin 2cos 7s t t t= + − determine all the points

where the object is not changing. 20. The position of an object is given by ( ) ( )8 10sins t t t= + determine where in the interval

[ ]0,12 the object is moving to the right and moving to the left.

21. Where in the range [ ]6,6− is the function ( ) ( )3 8cosf z z z= − is increasing and

decreasing. 22. Where in the range [ ]3,5− is the function ( ) ( ) ( )7cos sin 3R w w w= − + is increasing and

decreasing. 23. Where in the range [ ]0,10 is the function ( ) ( )9 15sinh t t= + is increasing and decreasing.

24. Using the definition of the derivative prove that ( )( ) ( )cos sind x xdx

= − .

25. Prove that ( )( ) ( ) ( )sec sec tand x x xdx

= .

26. Prove that ( )( ) ( )2cot cscd x xdx

= − .

27. Prove that ( )( ) ( ) ( )csc csc cotd x x xdx

= − .


Calculus I


Derivatives of Exponential and Logarithm Functions For problems 1 – 12 differentiate the given function. 1. ( ) 10 9z zg z = −

2. ( ) ( ) ( )4 119log 12logf x x x= +

3. ( ) 6 4t th t = − e

4. ( ) ( ) ( )12320ln logR x x x= +

5. ( ) ( )2 6 3 tQ t t t= − + e

6. 8 9v vy v= + 7. ( ) ( ) ( )6

4log lnU z z z z= −

8. ( ) ( ) ( )3log logh x x x=

9. ( ) 11 7

w

wf w −=

+ee

10. ( ) ( )3

1 4ln5

tf t

t+

=

11. ( ) ( )27log

7rr r

g r+

=

12. ( ) ( )4

ln

ttV tt

=e

13. Find the tangent line to ( ) ( )1 8 xf x x= − e at 1x = − .

14. Find the tangent line to ( ) ( )23 lnf x x x= at 1x = .

15. Find the tangent line to ( ) ( )3 8lnxf x x= +e at 2x = .


Calculus I


16. Determine if ( ) 4 3y yU y = − e is increasing or decreasing at the following points.

(a) 2y = − (b) 0y = (c) 3y =

17. Determine if ( ) ( )2

lnzy z

z= is increasing or decreasing at the following points.

(a) 12

z = (b) 2z = (c) 6z =

18. Determine if ( ) 2 xh x x= e is increasing or decreasing at the following points.

(a) 1x = − (b) 0x = (c) 2x =

Derivatives of Inverse Trig Functions For each of the following problems differentiate the given function. 1. ( ) ( ) ( )1sin 9sinf x x x−= +

2. ( ) ( ) ( )1 15sin cosC t t t− −= −

3. ( ) ( ) ( )1 1tan 4cosg z z z− −= +

4. ( ) ( ) ( )1 3 1sec cosh t t t t− −= −

5. ( ) ( ) ( )2 1sinf w w w w−= −

6. ( )( ) ( )( )1 1cot 1 cscy x x x− −= − +

7. ( ) ( )1

1tan

zQ zz−

+=

8. ( ) ( )( )

1

1

1 sin1 cos

tA t

t

−

−

+=

−


Calculus I


Derivatives of Hyperbolic Functions For each of the following problems differentiate the given function. 1. ( ) ( )2 3sinhh w w w= −

2. ( ) ( ) ( )cos coshg x x x= +

3. ( ) ( ) ( )3csch 7sinhH t t t= +

4. ( ) ( ) ( )tan tanhA r r r=

5. ( ) ( )coshxf x x= e

6. ( ) ( )sech 11

zf z

z+

=−

7. ( ) ( )( )

cothsinh

wQ w

w w=

+

Chain Rule

For problems 1 – 46 differentiate the given function.

1. ( ) ( )113 8g x x= −

2. ( ) 7 39g z z=

3. ( ) ( )639 2h t t t= + −

4. 3 28y w w= +

5. ( ) ( ) 2214 3R v v v−

= −

6. ( )( )8

26 5

H ww

=−


Calculus I


7. ( ) ( )4sin 4 7f x x x= +

8. ( ) ( )tan 1 2 xT x = − e

9. ( ) ( )( )2cos sing z z z= +

10. ( ) ( )2sech u u u= −

11. ( )( )cot 1 coty x= +

12. ( ) 21 tf t −= e

13. ( ) 612z zJ z −= e

14. ( ) ( )lnz zf z += e

15. ( ) ( )cos7 xB x =

16. 2 93x xz −=

17. ( ) ( )ln 6 zR z z= + e

18. ( ) ( )7 5 3lnh w w w w w= − + −

19. ( ) ( )( )ln 1 cscg t t= −

20. ( ) ( )1tan 3 2f v v−= −

21. ( ) ( )1sin 9h t t−=

22. ( ) ( ) ( )6cos 1 sinA t t t= − −

23. ( ) ( ) ( )ln 6 4secH z z z= −


Calculus I


24. ( ) ( ) ( )4 4tan tanf x x x= +

25. ( ) 24 86 7 uu u uf u − −= − +e e e

26. ( ) ( ) ( )8 8sec secg z z z= +

27. ( ) ( )54 1 2 9k w w w= − + +

28. ( ) ( ) 73 2 5 1 9 4h x x x x −= − + + +

29. ( ) ( ) ( )5 432 1 5 3T x x x= − −

30. ( ) ( )2 4 sin 1 2w z z z= + −

31. ( ) ( )8 4cosY t t t=

32. ( ) ( )46 ln 10 3f x x x= − +

33. ( ) ( ) ( )2sec 4 tanA z z z=

34. ( ) ( )4 6 95 ln vh v v v += + e

35. ( )2 8

4 7

x xf x

x

+

=+

e

36. ( ) ( )( )

3

62

4 1xg x

x x

+=

−

37. ( ) ( )csc 11 t

tg t −

−=

+ e

38. ( ) ( )( )

2

2

sin1 cos

zV z

z=

+


Calculus I


39. ( ) ( )( )ln coswU w w= e

40. ( ) ( ) ( )( )2tan 5 lnh t t t= −

41. 2

3ln2

xzx

+ = −

42. ( )7 2

vg v

v=

+e

43. ( ) 2 1 4f x x x= + +

44. ( )( )56 cos 8u w= +

45. ( ) ( )2 42 57 z zh z z z

−+= − + e

46. ( ) ( )( )3 2ln 7 sinA y y y= +

47. ( ) ( )6csc 8g x x=

48. ( ) ( ) ( )24 cos 9 ln 6 5V w w w= − + +

49. ( ) ( )3 6sin th t t −= e

50. ( ) ( ) ( )( )8sin sinr rB r = −e e

51. ( ) ( )( )2 2cos 1 cosf z z= +

52. Find the tangent line to ( ) ( )522 4f x x= − at 0x = .

53. Find the tangent line to ( ) ( )2 4 28ln 3xf x x+= − −e at 2x = − .


Calculus I


54. Determine where ( ) ( )43 9A t t t= − is increasing and decreasing.

55. Is ( ) ( ) ( )4 52 1 2h x x x= + − increasing or decreasing more in the interval [ ]2,3− ?

56. Determine where ( ) 3cos 32wU w w = + −

is increasing and decreasing in the interval

[ ]10,10− .

57. If the position of an object is given by ( ) ( )4sin 3 10 7s t t= − + . Determine where, if

anywhere, the object is not moving in the interval [ ]0,4 .

58. Determine where ( ) ( ) ( )6sin 2 7cos 3 3f x x x= − − is increasing and decreasing in the

interval [ ]3,2− .

59. Determine where ( ) ( ) 22 21 wH w w −= − e is increasing and decreasing.

60. What percentage of [ ]3,5− is the function ( ) 2 28 1 23z zg z − −= +e e decreasing?

61. The position of an object is given by ( ) ( )3 2ln 2 21 36 200s t t t t= − + + . During the first 10

hours of motion (assuming the motion starts at 0t = ) what percentage of the time is the object moving to the right?

62. For the function ( ) ( )21 ln 2 92xf x x x= − − + − determine each of the following.

(a) The interval on which the function is defined. (b) Where the function is increasing and decreasing.

Implicit Differentiation For problems 1 – 6 do each of the following.


1. 2 9 2x y =


Calculus I


2. 7

6 4xy

=

3. 4 31 5x y= + 4. 28 3x y− = 5. 2 24 6x y xy− = 6. ( )ln x y x=

For problems 7 – 21 find y′ by implicit differentiation. 7. 2 312 8y x y− = 8. 7 10 2 33 6 2y x y x−+ = − + 9. 3 1 14 8y x y− − −+ = 10. 4 6 3 310 7 4x y y x− −− = + 11. ( ) ( ) 4sin cos yx y+ = e

12. ( ) ( )ln secx y y+ =

13. ( )2 2 74 9y x y x− = +

14. 2 3 26 4 0x x y x− − + = 15. 4 3 38 2xy x y x−+ = 16. ( ) ( )3 cos sin 7yx x y x− =

17. ( ) ( )cos sin 9x y xy+ =e

18. 2 3 22x x y y+ + =


Calculus I


19. ( ) 1tan 3 7 6 4x y x−+ = −

20. 2 2 2 2

1x y x y+ = +e e

21. 3 4sin 2x x yy

+ = −

For problems 22 - 24 find the equation of the tangent line at the given point. 22. 23 1x y x+ = + at ( )4,3−

23. 2 2 6x y y x= − at ( )2,6

24. ( ) ( )2sin cos 1x y = at , 06π

For problems 25 – 27 determine if the function is increasing, decreasing or not changing at the given point. 25. 2 3 4 9x y y− = + at ( )2, 1−

26. 21 3x y x y− = +e e at ( )1,0

27. ( ) ( )2sin cosx y x yπ − + = at ,12π

For problems 28 - 31 assume that ( )x x t= , ( )y y t= and ( )z z t= and differentiate the given

equation with respect to t. 28. 4 26 3x z y− = − 29. 4 2 3x y y z=

30. ( )107 6 2 48yz y x z−= − +e

31. ( )2 3 2 2cos 0z x y x+ + =


Calculus I


Related Rates 1. In the following assume that x and y are both functions of t. Given 3x = , 2y = and 7y′ = determine x′ for the following equation. 3 4 2 7x y x y− = − 2. In the following assume that x and y are both functions of t. Given 6x π= , 4y = − and

12x′ = determine x′ for the following equation.

( ) ( )2 2 16 6 cos 2 1x y x y− − = + 3. In the following assume that x, y and z are all functions of t. Given 1x = − , 8y = , 2z = ,

4x′ = − and 7y′ = determine z′ for the following equation.

4 2 22 3yx x zz

+ = −

4. In the following assume that x, y and z are all functions of t. Given 2x = − , 3y = , 4z = ,

6y′ = and 0z′ = determine x′ for the following equation.

2 2 3 4 8x y z x z y= − − 5. The sides of a square are increasing at a rate of 10 cm/sec. How fast is the area enclosed by the square increasing when the area is 150 cm2. 6. The sides of an equilateral triangle are decreasing at a rate of 3 in/hr. How fast is the area enclosed by the triangle decreasing when the sides are 2 feet long? 7. A spherical balloon is being filled in such a way that the surface area is increasing at a rate of 20 cm2/sec when the radius is 2 meters. At what rate is air being pumped in the balloon when the radius is 2 meters? 8. A cylindrical tank of radius 2.5 feet is being drained of water at a rate of 0.25 ft3/sec. How fast is the height of the water decreasing? 9. A hot air balloon is attached to a spool of rope that is 125 feet away from the balloon when it is on the ground. The hot air balloon rises straight up in such a way that the length of rope increases at a rate of 15 ft/sec. How fast is the hot air balloon rising 20 seconds after is lifts off? See the (probably bad) sketch below to help visualize the problem.


Calculus I


10. A rock is dropped straight off a bridge that is 50 meters above the ground. Another person is 7 meters away on the same bridge. At what rate is the distance between the rock and the second person increasing just as the rock hits the ground? 11. A person is 8 meters away from a road and there is a car that is initially 800 meters away approaching the person at a speed of 45 m/sec. At what rate is the distance between the person and the car changing (a) 5 seconds after the start, (b) when the car is directly in front of the person and (c) 10 seconds after the car has passed the person. See the (probably bad) sketch below to help visualize the problem.

12. Two cars are initially 1200 miles apart. At the same time Car A starts driving at 35 mph to the east while Car B starts driving at 55 mph to the north (see sketch below for this initial setup). At what rate is the distance between the two cars changing after (a) 2 hours of travel, (b) 20 hours of travel and (c) 40 hours of travel?

13. Repeat problem 12 above except for this problem assume that Car A starts traveling 4 hours after Car B starts traveling. For parts (a), (b) and (c) assume that these are travel times for Car B. 14. Two people are on a city block. See the sketch below for placement and distances. Person A is on the northeast corner and Person B is on the southwest corner. Person A starts walking towards the southeast corner at a rate of 3 ft/sec. Four seconds later Person B starts walking


Calculus I


towards the southeast corner at a rate of 2 ft/sec. At what rate is the distance between them changing (a) 3 seconds after Person A starts walking and (b) after Person A has covered half the distance?

15. A person is standing 75 meters away from a kite and has a spool of string attached to the kite. The kite starts to rise straight up in the air at a rate of 2 m/sec and at the same time the person starts to move towards the kites launch point at a rate of 0.75 m/sec. Is the length string increasing or decreasing after (a) 4 seconds and (b) 20 seconds. 16. A person lights the fuse on a model rocket and starts to move away from the rocket at a rate of 3 ft/sec. Five seconds after lighting the fuse the rocket launches straight up into the air at a rate of 10 ft/sec. Is the distance between the person and the rocket increasing or decreasing (a) 5 seconds after launch and (b) 10 seconds after launch? 17. A light is suspended above the ground and is being lowered towards the ground at a rate of 9 in/sec. A 6 foot tall person is on the ground and 8 feet away from the light. At what rate is the persons shadow increasing then the light is 15 feet above the ground? 18. A light is fixed on a wall 10 meters above the floor. Twelve meters away from the wall a pole is being raised straight up at a rate of 45 cm/sec. When the pole is 6 meters tall at what rate is the tip of the shadow moving (a) away from the pole and (b) away from the wall? 19. A light is on the top of a 15 foot tall pole. A 5 foot tall person moves away from the pole at a rate of 2.5 ft/sec. After moving for 8 seconds at what rate is the tip of the shadow moving (a) away from the person and (b) away from the pole? 20. A tank of water is in the shape of a cone and is leaking water at a rate of 35 cm3/sec. The base radius of the tank is 1 meter and the height of the tank is 2.5 meters. When the depth of the water is 1.25 meters at what rate is the (a) depth changing and (b) the radius of the top of the water changing? 21. A trough of water is 20 meters long and its ends are in the shape of an isosceles triangle whose width is 7 meters and height is 10 meters. Assume that the two equal length sides of the triangle are the sides of the water tank and the other side of the triangle is the top of the tank and is parallel to the ground. Water is being pumped into the tank at a rate of 2 m3/min. When the


Calculus I


water is 6 meters deep at what rate is (a) depth changing and (b) the width of the top of the water changing? 22. A trough of water is 9 feet long and its ends are in the shape of an equilateral triangle whose sides are 1.5 feet long. Assume that the top of the tank is parallel to the ground. If water is being pumped out of the tank at what rate is the depth of the water changing when the depth is 0.75 feet? 23. The angle of elevation (depression) is the angle formed by a horizontal line and a line joining the observer’s eye to an object above (below) the horizontal line. Two people are on the roof of buildings separated by at 25 foot wide road. Person A is 100 feet above Person B and drops a rock off the roof of their building and it falls at a rate of 3 ft/sec. (a) At what rate is the angle of elevation changing as Person B watches the rock fall when the rock is 25 feet above Person B? (b) At what rate is the angle of depression changing as Person B watches the rock fall when the rock is 65 feet below Person B? 24. The angle of elevation is the angle formed by a horizontal line and a line joining the observer’s eye to an object above the horizontal line. A person is standing 15 meters away from a building and watching an outside elevator move down the face of the building. When the angle of elevation is 1 radians it is changing at a rate of 0.15 radians/sec. At this point in time what is the speed of the elevator? 25. The angle of elevation is the angle formed by a horizontal line and a line joining the observer’s eye to an object above the horizontal line. A person is 24 feet away from a building and watching an outside elevator move up the face of the building. The elevator is moving up at a rate of 4 ft/sec and the person is moving towards the building at a rate of 0.75 ft/sec. Assuming that the elevator started moving from the ground at the same time that the person started walking is the angle of elevation increasing or decreasing after 10 seconds?

Higher Order Derivatives For problems 1 – 9 determine the fourth derivative of the given function. 1. ( ) 8 6 4 22 7 20 3f z z z z z= + − + −

2. 4 3 26 5 4 3 2y t t t t= − + − + 3. ( ) 2 3 46 7V t t t t− − −= + −


Calculus I


4. ( ) 3 5

3 1 34 2

g xx x x

= − +

5. ( ) 9438 5h x x x x= − +

6. ( ) 2354

32 13

h y yy y

= − +

7. ( ) ( ) ( )2

39sin sin 4 7cos zy z z= − +

8. ( ) ( )1 82 3 9 ln 6x xR x x− += − +e e

9. ( ) ( ) ( ) ( )6 7ln cos 4 9sin 2 tf t t t t= − + + e

For problems 10 – 20 determine the second derivative of the given function.

10. ( ) ( )2cos 2 7Q w w= −

11. ( ) ( )2sin 1 xf z = + e

12. ( )tan 3y x=

13. ( )csc 8z w=

14. ( ) 24 9u uf u += e

15. ( ) ( )2ln 3h x x x= −

16. ( ) ( )( )ln 3 cosg z z= +

17. ( )4

16

f xx x

=+

18. ( ) ( ) ( ) 33sin 8cos 2f x x x

−= +


Calculus I


19. ( ) ( )3sin 2f t t=

20. ( ) ( )4tanA w w=

For problems 21 – 23 determine the third derivative of the given function. 21. ( ) ( )sec 3g x x=

22. 31 2ty −= e

23. ( ) ( )2cosh w w w= −

For problems 24 - 27 determine the second derivative of the given function. 24. 2 46 3 9y y x x− = + 25. 3 2 24 11 2y x x y− = − 26. 34 1y x y+ = −e 27. ( ) 2cos 3 4y x y= +

Logarithmic Differentiation For problems 1 – 6 use logarithmic differentiation to find the first derivative of the given function.

1. ( ) ( )( )48 2cos 3 6 3h x x x x= +

2. ( ) 52 3 54 2 9 7 2f w w w w w w= + − + +

3. ( ) ( )( )

32

42

1 7

2 3 4

zh z

z z

+=

+ +


Calculus I


4. ( ) ( )( )

1 sin 22 tan

xg x

x x+

=−

5. ( ) ( )( )

10

2

9 3sin 7

th t

t t−

=

6. ( )

( )( )4 72 2

cos 13 8

1 2 5

xxyx x x

−+=

+ +

For problems 6 – 9 find the first derivative of the given function.

6. ( )ln xy x=

7. ( ) ( ) 6sin 4

tR t t=

8. ( ) ( )32 826

w wh w w

+ += −

9. ( ) [ ]212 3 zg z z z −= +


CALCULUS I Applications of Derivatives

Paul Dawkins

Calculus I


Table of Contents Preface ............................................................................................................................................. ii Applications of Derivatives ........................................................................................................... 2

Introduction ................................................................................................................................................ 3 Rates of Change ......................................................................................................................................... 5 Critical Points ............................................................................................................................................. 8 Minimum and Maximum Values ............................................................................................................. 14 Finding Absolute Extrema ....................................................................................................................... 22 The Shape of a Graph, Part I .................................................................................................................... 28 The Shape of a Graph, Part II ................................................................................................................... 37 The Mean Value Theorem ........................................................................................................................ 46 Optimization ............................................................................................................................................. 53 More Optimization Problems ................................................................................................................... 67 Indeterminate Forms and L’Hospital’s Rule ............................................................................................ 82 Linear Approximations ............................................................................................................................ 89 Differentials.............................................................................................................................................. 92 Newton’s Method ..................................................................................................................................... 95 Business Applications ............................................................................................................................ 100


Calculus I

© 2007 Paul Dawkins ii http://tutorial.math.lamar.edu/terms.aspx

Preface Here are my online notes for my Calculus I course that I teach here at Lamar University. Despite the fact that these are my “class notes”, they should be accessible to anyone wanting to learn Calculus I or needing a refresher in some of the early topics in calculus. I’ve tried to make these notes as self contained as possible and so all the information needed to read through them is either from an Algebra or Trig class or contained in other sections of the notes. Here are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed.

1. Because I wanted to make this a fairly complete set of notes for anyone wanting to learn calculus I have included some material that I do not usually have time to cover in class and because this changes from semester to semester it is not noted here. You will need to find one of your fellow class mates to see if there is something in these notes that wasn’t covered in class.

2. Because I want these notes to provide some more examples for you to read through, I don’t always work the same problems in class as those given in the notes. Likewise, even if I do work some of the problems in here I may work fewer problems in class than are presented here.

3. Sometimes questions in class will lead down paths that are not covered here. I try to anticipate as many of the questions as possible when writing these up, but the reality is that I can’t anticipate all the questions. Sometimes a very good question gets asked in class that leads to insights that I’ve not included here. You should always talk to someone who was in class on the day you missed and compare these notes to their notes and see what the differences are.

4. This is somewhat related to the previous three items, but is important enough to merit its own item. THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!! Using these notes as a substitute for class is liable to get you in trouble. As already noted not everything in these notes is covered in class and often material or insights not in these notes is covered in class.

Applications of Derivatives


Calculus I


Introduction In the previous chapter we focused almost exclusively on the computation of derivatives. In this chapter will focus on applications of derivatives. It is important to always remember that we didn’t spend a whole chapter talking about computing derivatives just to be talking about them. There are many very important applications to derivatives. The two main applications that we’ll be looking at in this chapter are using derivatives to determine information about graphs of functions and optimization problems. These will not be the only applications however. We will be revisiting limits and taking a look at an application of derivatives that will allow us to compute limits that we haven’t been able to compute previously. We will also see how derivatives can be used to estimate solutions to equations. Here is a listing of the topics in this section. Rates of Change – The point of this section is to remind us of the application/interpretation of derivatives that we were dealing with in the previous chapter. Namely, rates of change. Critical Points – In this section we will define critical points. Critical points will show up in many of the sections in this chapter so it will be important to understand them. Minimum and Maximum Values – In this section we will take a look at some of the basic definitions and facts involving minimum and maximum values of functions. Finding Absolute Extrema – Here is the first application of derivatives that we’ll look at in this chapter. We will be determining the largest and smallest value of a function on an interval. The Shape of a Graph, Part I – We will start looking at the information that the first derivatives can tell us about the graph of a function. We will be looking at increasing/decreasing functions as well as the First Derivative Test. The Shape of a Graph, Part II – In this section we will look at the information about the graph of a function that the second derivatives can tell us. We will look at inflection points, concavity, and the Second Derivative Test. The Mean Value Theorem – Here we will take a look that the Mean Value Theorem. Optimization Problems – This is the second major application of derivatives in this chapter. In this section we will look at optimizing a function, possible subject to some constraint. More Optimization Problems – Here are even more optimization problems. L’Hospital’s Rule and Indeterminate Forms – This isn’t the first time that we’ve looked at indeterminate forms. In this section we will take a look at L’Hospital’s Rule. This rule will allow us to compute some limits that we couldn’t do until this section.


Calculus I


Linear Approximations – Here we will use derivatives to compute a linear approximation to a function. As we will see however, we’ve actually already done this. Differentials – We will look at differentials in this section as well as an application for them. Newton’s Method – With this application of derivatives we’ll see how to approximate solutions to an equation. Business Applications – Here we will take a quick look at some applications of derivatives to the business field.


Calculus I


Rates of Change The purpose of this section is to remind us of one of the more important applications of derivatives. That is the fact that ( )f x′ represents the rate of change of ( )f x . This is an

application that we repeatedly saw in the previous chapter. Almost every section in the previous chapter contained at least one problem dealing with this application of derivatives. While this application will arise occasionally in this chapter we are going to focus more on other applications in this chapter. So, to make sure that we don’t forget about this application here is a brief set of examples concentrating on the rate of change application of derivatives. Note that the point of these examples is to remind you of material covered in the previous chapter and not to teach you how to do these kinds of problems. If you don’t recall how to do these kinds of examples you’ll need to go back and review the previous chapter. Example 1 Determine all the points where the following function is not changing. ( ) ( )5 6 10cos 2g x x x= − − Solution First we’ll need to take the derivative of the function. ( ) ( )6 20sin 2g x x′ = − +

Now, the function will not be changing if the rate of change is zero and so to answer this question we need to determine where the derivative is zero. So, let’s set this equal to zero and solve.

( ) ( )6 20sin 2 0 sin 2 0.320

x x 6− + = ⇒ = =

The solution to this is then,

2 0.3047 2 OR 2 2.8369 2 0, 1, 2,

0.1524 OR 1.4185 0, 1, 2,x n x n nx n x n n

π ππ π

= + = + = ± ±= + = + = ± ±

……

If you don’t recall how to solve trig equations check out the Solving Trig Equations sections in the Review Chapter. Example 2 Determine where the following function is increasing and decreasing. ( ) 5 4 327 45 130 150A t t t t= − − + Solution As with the first problem we first need to take the derivative of the function.

( ) ( )4 3 2 2 2135 180 390 15 9 12 26A t t t t t t t′ = − − = − −

Next, we need to determine where the function isn’t changing. This is at,


Calculus I


( ) ( )0

12 144 4 9 26 12 1080 12 6 30 2 30 1.159, 2.49218 18 18 3

t

t

=

± − − ± ± ±= = = = = −

So, the function is not changing at three values of t. Finally, to determine where the function is increasing or decreasing we need to determine where the derivative is positive or negative. Recall that if the derivative is positive then the function must be increasing and if the derivative is negative then the function must be decreasing. The following number line gives this information.

So, from this number line we can see that we have the following increasing and decreasing information. Increasing : 1.159, 2.492 Decreasing : 1.159 0, 0 2.492t t t t− ∞ < < − < < ∞ − < < < < If you don’t remember how to solve polynomial and rational inequalities then you should check out the appropriate sections in the Review Chapter. Finally, we can’t forget about Related Rates problems. Example 3 Two cars start out 500 miles apart. Car A is to the west of Car B and starts driving to the east (i.e. towards Car B) at 35 mph and at the same time Car B starts driving south at 50 mph. After 3 hours of driving at what rate is the distance between the two cars changing? Is it increasing or decreasing? Solution The first thing to do here is to get sketch a figure showing the situation.

In this figure y represents the distance driven by Car B and x represents the distance separating Car A from Car B’s initial position and z represents the distance separating the two cars. After 3


Calculus I


hours driving time with have the following values of x and y. ( ) ( )500 35 3 395 50 3 150x y= − = = =

We can use the Pythagorean theorem to find z at this time as follows,

2 2 2395 150 178525 178525 422.5222z z= + = ⇒ = = Now, to answer this question we will need to determine z′ given that 35x′ = − and 50y′ = . Do you agree with the signs on the two given rates? Remember that a rate is negative if the quantity is decreasing and positive if the quantity is increasing. We can again use the Pythagorean theorem here. First, write it down and the remember that x, y, and z are all changing with time and so differentiate the equation using Implicit Differentiation. 2 2 2 2 2 2z x y zz xx yy′ ′ ′= + ⇒ = + Finally, all we need to do is cancel a two from everything, plug in for the known quantities and solve for z′ .

( ) ( ) ( ) ( )( ) 6325422.5222 395 35 150 50 14.9696422.5222

z z −′ ′= − + ⇒ = = −

So, after three hours the distance between them is decreasing at a rate of 14.9696 mph. So, in this section we covered three “standard” problems using the idea that the derivative of a function gives its rate of change. As mentioned earlier, this chapter will be focusing more on other applications than the idea of rate of change, however, we can’t forget this application as it is a very important one.


Calculus I


Critical Points Critical points will show up throughout a majority of this chapter so we first need to define them and work a few examples before getting into the sections that actually use them. Definition We say that x c= is a critical point of the function f(x) if ( )f c exists and if either of the

following are true. ( ) ( )0 OR doesn't existf c f c′ ′= Note that we require that ( )f c exists in order for x c= to actually be a critical point. This is an

important, and often overlooked, point. The main point of this section is to work some examples finding critical points. So, let’s work some examples. Example 1 Determine all the critical points for the function. ( ) 5 4 36 33 30 100f x x x x= + − + Solution We first need the derivative of the function in order to find the critical points and so let’s get that and notice that we’ll factor it as much as possible to make our life easier when we go to find the critical points.

( )( )( ) ( )

4 3 2

2 2

2

30 132 90

6 5 22 15

6 5 3 5

f x x x x

x x x

x x x

′ = + −

= + −

= − +

Now, our derivative is a polynomial and so will exist everywhere. Therefore the only critical points will be those values of x which make the derivative zero. So, we must solve. ( )( )26 5 3 5 0x x x− + = Because this is the factored form of the derivative it’s pretty easy to identify the three critical points. They are,

35, 0,5

x x x= − = =

Polynomials are usually fairly simple functions to find critical points for provided the degree doesn’t get so large that we have trouble finding the roots of the derivative.


Calculus I


Most of the more “interesting” functions for finding critical points aren’t polynomials however. So let’s take a look at some functions that require a little more effort on our part. Example 2 Determine all the critical points for the function.

( ) ( )3 2 2 1g t t t= − Solution To find the derivative it’s probably easiest to do a little simplification before we actually differentiate. Let’s multiply the root through the parenthesis and simplify as much as possible. This will allow us to avoid using the product rule when taking the derivative.

( ) ( )2 5 23 3 32 1 2g t t t t t= − = −

Now differentiate.

( )2

2 1 33 3

13

10 2 10 23 3 3 3

tg t t tt

−′ = − = −

We will need to be careful with this problem. When faced with a negative exponent it is often best to eliminate the minus sign in the exponent as we did above. This isn’t really required but it can make our life easier on occasion if we do that. Notice as well that eliminating the negative exponent in the second term allows us to correctly identify why 0t = is a critical point for this function. Once we move the second term to the denominator we can clearly see that the derivative doesn’t exist at 0t = and so this will be a critical point. If you don’t get rid of the negative exponent in the second term many people will incorrectly state that 0t = is a critical point because the derivative is zero at 0t = . While this may seem like a silly point, after all in each case 0t = is identified as a critical point, it is sometimes important to know why a point is a critical point. In fact, in a couple of sections we’ll see a fact that only works for critical points in which the derivative is zero. So, we’ve found one critical point (where the derivative doesn’t exist), but we now need to determine where the derivative is zero (provided it is of course…). To help with this it’s usually best to combine the two terms into a single rational expression. So, getting a common denominator and combining gives us,

( ) 13

10 2

3

tg tt

−′ =

Notice that we still have 0t = as a critical point. Doing this kind of combining should never lose critical points, it’s only being done to help us find them. As we can see it’s now become much easier to quickly determine where the derivative will be zero. Recall that a rational expression will only be zero if its numerator is zero (and provided the denominator isn’t also zero at that


Calculus I


point of course).

So, in this case we can see that the numerator will be zero if 15

t = and so there are two critical

points for this function.

10 and5

t t= =

Example 3 Determine all the critical points for the function.

( )2

2

16

wR ww w

+=

− −

Solution We’ll leave it to you to verify that using the quotient rule we get that the derivative is,

( )( ) ( )

2 2

2 22 2

14 1 14 1

6 6

w w w wR ww w w w

− − + + −′ = = −− − − −

Notice that we factored a “-1” out of the numerator to help a little with finding the critical points. This negative out in front will not affect the derivative whether or not the derivative is zero or not exist but will make our work a little easier. Now, we have two issues to deal with. First the derivative will not exist if there is division by zero in the denominator. So we need to solve, ( ) ( )2 6 3 2 0w w w w− − = − + = We didn’t bother squaring this since if this is zero, then zero squared is still zero and if it isn’t zero then squaring it won’t make it zero. So, we can see from this that the derivative will not exist at 3w = and 2w = − . However, these are NOT critical points since the function will also not exist at these points. Recall that in order for a point to be a critical point the function must actually exist at that point. At this point we need to be careful. The numerator doesn’t factor, but that doesn’t mean that there aren’t any critical points where the derivative is zero. We can use the quadratic formula on the numerator to determine if the fraction as a whole is ever zero.

( ) ( ) ( )

( )

214 14 4 1 1 14 200 14 10 2 7 5 22 1 2 2

w− ± − − − ± − ±

= = = = − ±

So, we get two critical points. Also, these are not “nice” integers or fractions. This will happen on occasion. Don’t get too locked into answers always being “nice”. Often they aren’t. Note as well that we only use real numbers for critical points. So, if upon solving the quadratic in the numerator, we had gotten complex number these would not have been considered critical points.


Calculus I


Summarizing, we have two critical points. They are,

7 5 2, 7 5 2w w= − + = − − Again, remember that while the derivative doesn’t exist at 3w = and 2w = − neither does the function and so these two points are not critical points for this function. So far all the examples have not had any trig functions, exponential functions, etc. in them. We shouldn’t expect that to always be the case. So, let’s take a look at some examples that don’t just involve powers of x. Example 4 Determine all the critical points for the function. ( )6 4cos 3y x x= − Solution First get the derivative and don’t forget to use the chain rule on the second term. ( )6 12sin 3y x′ = + Now, this will exist everywhere and so there won’t be any critical points for which the derivative doesn’t exist. The only critical points will come from points that make the derivative zero. We will need to solve,

( )

( )

6 12sin 3 01sin 32

x

x

+ =

= −

Solving this equation gives the following.

3 3.6652 2 , 0, 1, 2,3 5.7596 2 , 0, 1, 2,

x n nx n n

ππ

= + = ± ±= + = ± ±

……

Don’t forget the 2π n on these! There will be problems down the road in which we will miss solutions without this! Also make sure that it gets put on at this stage! Now divide by 3 to get all the critical points for this function.

21.2217 , 0, 1, 2,3

21.9199 , 0, 1, 2,3

nx n

nx n

π

π

= + = ± ±

= + = ± ±

…

…

Notice that in the previous example we got an infinite number of critical points. That will happen on occasion so don’t worry about it when it happens.


Calculus I


Example 5 Determine all the critical points for the function.

( ) 2310 th t t −= e Solution Here’s the derivative for this function.

( ) ( )2 2 2 23 3 3 2 310 10 2 10 20t t t th t t t t− − − −′ = + − = −e e e e Now, this looks unpleasant, however with a little factoring we can clean things up a little as follows,

( ) ( )23 210 1 2th t t−′ = −e This function will exist everywhere and so no critical points will come from that. Determining where this is zero is easier than it looks. We know that exponentials are never zero and so the only way the derivative will be zero is if,

2

2

2

1 2 01 212

tt

t

− =

=

=

We will have two critical points for this function.

12

t = ±

Example 6 Determine all the critical points for the function. ( ) ( )2 ln 3 6f x x x= + Solution Before getting the derivative let’s notice that since we can’t take the log of a negative number or zero we will only be able to look at 0x > . The derivative is then,

( ) ( )

( )( )( )

2 32 ln 33

2 ln 3

2ln 3 1

f x x x xx

x x x

x x

′ = +

= +

= +

Now, this derivative will not exist if x is a negative number or if 0x = , but then again neither will the function and so these are not critical points. Remember that the function will only exist if

0x > and nicely enough the derivative will also only exist if 0x > and so the only thing we


Calculus I


need to worry about is where the derivative is zero. First note that, despite appearances, the derivative will not be zero for 0x = . As noted above the derivative doesn’t exist at 0x = because of the natural logarithm and so the derivative can’t be zero there! So, the derivative will only be zero if,

( )

( )

2ln 3 1 01ln 32

x

x

+ =

= −

Recall that we can solve this by exponentiating both sides.

( )1

ln 3 2

12

12

3

1 13 3

x

x

x

−

−

−

=

=

= =

e e

e

ee

There is a single critical point for this function. Let’s work one more problem to make a point. Example 7 Determine all the critical points for the function. ( ) 2xf x x= e Solution Note that this function is not much different from the function used in Example 5. In this case the derivative is,

( ) ( ) ( )2 2 2 22 1 2x x xf x x x x′ = + = +e e e This function will never be zero for any real value of x. The exponential is never zero of course and the polynomial will only be zero if x is complex and recall that we only want real values of x for critical points. Therefore, this function will not have any critical points. It is important to note that not all functions will have critical points! In this course most of the functions that we will be looking at do have critical points. That is only because those problems make for more interesting examples. Do not let this fact lead you to always expect that a function will have critical points. Sometimes they don’t as this final example has shown.


Calculus I


Minimum and Maximum Values Many of our applications in this chapter will revolve around minimum and maximum values of a function. While we can all visualize the minimum and maximum values of a function we want to be a little more specific in our work here. In particular we want to differentiate between two types of minimum or maximum values. The following definition gives the types of minimums and/or maximums values that we’ll be looking at. Definition 1. We say that f(x) has an absolute (or global) maximum at x c= if ( ) ( )f x f c≤ for every x

in the domain we are working on. 2. We say that f(x) has a relative (or local) maximum at x c= if ( ) ( )f x f c≤ for every x in

some open interval around x c= . 3. We say that f(x) has an absolute (or global) minimum at x c= if ( ) ( )f x f c≥ for every x

in the domain we are working on. 4. We say that f(x) has a relative (or local) minimum at x c= if ( ) ( )f x f c≥ for every x in

some open interval around x c= . Note that when we say an “open interval around x c= ” we mean that we can find some interval

( ),a b , not including the endpoints, such that a c b< < . Or, in other words, c will be contained

somewhere inside the interval and will not be either of the endpoints. Also, we will collectively call the minimum and maximum points of a function the extrema of the function. So, relative extrema will refer to the relative minimums and maximums while absolute extrema refer to the absolute minimums and maximums. Now, let’s talk a little bit about the subtle difference between the absolute and relative in the definition above. We will have an absolute maximum (or minimum) at x c= provided f(c) is the largest (or smallest) value that the function will ever take on the domain that we are working on. Also, when we say the “domain we are working on” this simply means the range of x’s that we have chosen to work with for a given problem. There may be other values of x that we can actually plug into the function but have excluded them for some reason. A relative maximum or minimum is slightly different. All that’s required for a point to be a relative maximum or minimum is for that point to be a maximum or minimum in some interval of x’s around x c= . There may be larger or smaller values of the function at some other place, but relative to x c= , or local to x c= , f(c) is larger or smaller than all the other function values that are near it.


Calculus I


Note as well that in order for a point to be a relative extrema we must be able to look at function values on both sides of x c= to see if it really is a maximum or minimum at that point. This means that relative extrema do not occur at the end points of a domain. They can only occur interior to the domain. There is actually some debate on the preceding point. Some folks do feel that relative extrema can occur on the end points of a domain. However, in this class we will be using the definition that says that they can’t occur at the end points of a domain. It’s usually easier to get a feel for the definitions by taking a quick look at a graph.

For the function shown in this graph we have relative maximums at x b= and x d= . Both of these point are relative maximums since they are interior to the domain shown and are the largest point on the graph in some interval around the point. We also have a relative minimum at x c= since this point is interior to the domain and is the lowest point on the graph in an interval around it. The far right end point, x e= , will not be a relative minimum since it is an end point. The function will have an absolute maximum at x d= and an absolute minimum at x a= . These two points are the largest and smallest that the function will ever be. We can also notice that the absolute extrema for a function will occur at either the endpoints of the domain or at relative extrema. We will use this idea in later sections so it’s more important than it might seem at the present time. Let’s take a quick look at some examples to make sure that we have the definitions of absolute extrema and relative extrema straight. Example 1 Identify the absolute extrema and relative extrema for the following function. ( ) [ ]2 on 1,2f x x= − Solution Since this function is easy enough to graph let’s do that. However, we only want the graph on the interval [-1,2]. Here is the graph,


Calculus I


Note that we used dots at the end of the graph to remind us that the graph ends at these points. We can now identify the extrema from the graph. It looks like we’ve got a relative and absolute minimum of zero at 0x = and an absolute maximum of four at 2x = . Note that 1x = − is not a relative maximum since it is at the end point of the interval. This function doesn’t have any relative maximums. As we saw in the previous example functions do not have to have relative extrema. It is completely possible for a function to not have a relative maximum and/or a relative minimum. Example 2 Identify the absolute extrema and relative extrema for the following function. ( ) [ ]2 on 2, 2f x x= − Solution Here is the graph for this function.

In this case we still have a relative and absolute minimum of zero at 0x = . We also still have an absolute maximum of four. However, unlike the first example this will occur at two points,

2x = − and 2x = . Again, the function doesn’t have any relative maximums. As this example has shown there can only be a single absolute maximum or absolute minimum value, but they can occur at more than one place in the domain.


Calculus I


Example 3 Identify the absolute extrema and relative extrema for the following function. ( ) 2f x x= Solution In this case we’ve given no domain and so the assumption is that we will take the largest possible domain. For this function that means all the real numbers. Here is the graph.

In this case the graph doesn’t stop increasing at either end and so there are no maximums of any kind for this function. No matter which point we pick on the graph there will be points both larger and smaller than it on either side so we can’t have any maximums (or any kind, relative or absolute) in a graph. We still have a relative and absolute minimum value of zero at 0x = . So, some graphs can have minimums but not maximums. Likewise, a graph could have maximums but not minimums. Example 4 Identify the absolute extrema and relative extrema for the following function. ( ) [ ]3 on 2, 2f x x= − Solution Here is the graph for this function.


Calculus I


This function has an absolute maximum of eight at 2x = and an absolute minimum of negative eight at 2x = − . This function has no relative extrema. So, a function doesn’t have to have relative extrema as this example has shown. Example 5 Identify the absolute extrema and relative extrema for the following function. ( ) 3f x x= Solution Again, we aren’t restricting the domain this time so here’s the graph.

In this case the function has no relative extrema and no absolute extrema. As we’ve seen in the previous example functions don’t have to have any kind of extrema, relative or absolute. Example 6 Identify the absolute extrema and relative extrema for the following function. ( ) ( )cosf x x= Solution We’ve not restricted the domain for this function. Here is the graph.

Cosine has extrema (relative and absolute) that occur at many points. Cosine has both relative and absolute maximums of 1 at 4 , 2 , 0, 2 , 4 ,x π π π π= − −… … Cosine also has both relative and absolute minimums of -1 at 3 , , , 3 ,x π π π π= − −… …


Calculus I


As this example has shown a graph can in fact have extrema occurring at a large number (infinite in this case) of points. We’ve now worked quite a few examples and we can use these examples to see a nice fact about absolute extrema. First let’s notice that all the functions above were continuous functions. Next notice that every time we restricted the domain to a closed interval (i.e. the interval contains its end points) we got absolute maximums and absolute minimums. Finally, in only one of the three examples in which we did not restrict the domain did we get both an absolute maximum and an absolute minimum. These observations lead us the following theorem. Extreme Value Theorem Suppose that ( )f x is continuous on the interval [a,b] then there are two numbers ,a c d b≤ ≤ so

that ( )f c is an absolute maximum for the function and ( )f d is an absolute minimum for the

function. So, if we have a continuous function on an interval [a,b] then we are guaranteed to have both an absolute maximum and an absolute minimum for the function somewhere in the interval. The theorem doesn’t tell us where they will occur or if they will occur more than once, but at least it tells us that they do exist somewhere. Sometimes, all that we need to know is that they do exist. This theorem doesn’t say anything about absolute extrema if we aren’t working on an interval. We saw examples of functions above that had both absolute extrema, one absolute extrema, and no absolute extrema when we didn’t restrict ourselves down to an interval. The requirement that a function be continuous is also required in order for us to use the theorem. Consider the case of

( ) 2

1 on [ 1,1]f xx

= −

Here’s the graph.


Calculus I


This function is not continuous at 0x = as we move in towards zero the function is approaching infinity. So, the function does not have an absolute maximum. Note that it does have an absolute minimum however. In fact the absolute minimum occurs twice at both 1x = − and 1x = . If we changed the interval a little to say,

( ) 2

1 1on ,12

f xx

=

the function would now have both absolute extrema. We may only run into problems if the interval contains the point of discontinuity. If it doesn’t then the theorem will hold. We should also point out that just because a function is not continuous at a point that doesn’t mean that it won’t have both absolute extrema in an interval that contains that point. Below is the graph of a function that is not continuous at a point in the given interval and yet has both absolute extrema.

This graph is not continuous at x c= , yet it does have both an absolute maximum ( x b= ) and an absolute minimum ( x c= ). Also note that, in this case one of the absolute extrema occurred at the point of discontinuity, but it doesn’t need to. The absolute minimum could just have easily been at the other end point or at some other point interior to the region. The point here is that this graph is not continuous and yet does have both absolute extrema The point of all this is that we need to be careful to only use the Extreme Value Theorem when the conditions of the theorem are met and not misinterpret the results if the conditions aren’t met. In order to use the Extreme Value Theorem we must have an interval and the function must be continuous on that interval. If we don’t have an interval and/or the function isn’t continuous on the interval then the function may or may not have absolute extrema. We need to discuss one final topic in this section before moving on to the first major application of the derivative that we’re going to be looking at in this chapter.


Calculus I


Fermat’s Theorem If ( )f x has a relative extrema at x c= and ( )f c′ exists then x c= is a critical point of

( )f x . In fact, it will be a critical point such that ( ) 0f c′ = .

Note that we can say that ( ) 0f c′ = because we are also assuming that ( )f c′ exists.

This theorem tells us that there is a nice relationship between relative extrema and critical points. In fact it will allow us to get a list of all possible relative extrema. Since a relative extrema must be a critical point the list of all critical points will give us a list of all possible relative extrema. Consider the case of ( ) 2f x x= . We saw that this function had a relative minimum at 0x = in

several earlier examples. So according to Fermat’s theorem 0x = should be a critical point. The derivative of the function is, ( ) 2f x x′ = Sure enough 0x = is a critical point. Be careful not to misuse this theorem. It doesn’t say that a critical point will be a relative extrema. To see this, consider the following case. ( ) ( )3 23f x x f x x′= = Clearly 0x = is a critical point. However we saw in an earlier example this function has no relative extrema of any kind. So, critical points do not have to be relative extrema. Also note that this theorem says nothing about absolute extrema. An absolute extrema may or may not be a critical point. To see the proof of this theorem see the Proofs From Derivative Applications section of the Extras chapter.


Calculus I


Finding Absolute Extrema It’s now time to see our first major application of derivatives in this chapter. Given a continuous function, f(x), on an interval [a,b] we want to determine the absolute extrema of the function. To do this we will need many of the ideas that we looked at in the previous section. First, since we have an interval and we are assuming that the function is continuous the Extreme Value Theorem tells us that we can in fact do this. This is a good thing of course. We don’t want to be trying to find something that may not exist. Next, we saw in the previous section that absolute extrema can occur at endpoints or at relative extrema. Also, from Fermat’s Theorem we know that the list of critical points is also a list of all possible relative extrema. So the endpoints along with the list of all critical points will in fact be a list of all possible absolute extrema. Now we just need to recall that the absolute extrema are nothing more than the largest and smallest values that a function will take so all that we really need to do is get a list of possible absolute extrema, plug these points into our function and then identify the largest and smallest values. Here is the procedure for finding absolute extrema. Finding Absolute Extrema of f(x) on [a,b].

0. Verify that the function is continuous on the interval [a,b]. 1. Find all critical points of f(x) that are in the interval [a,b]. This makes sense if you think

about it. Since we are only interested in what the function is doing in this interval we don’t care about critical points that fall outside the interval.

2. Evaluate the function at the critical points found in step 1 and the end points. 3. Identify the absolute extrema.

There really isn’t a whole lot to this procedure. We called the first step in the process step 0, mostly because all of the functions that we’re going to look at here are going to be continuous, but it is something that we do need to be careful with. This process will only work if we have a function that is continuous on the given interval. The most labor intensive step of this process is the second step (step 1) where we find the critical points. It is also important to note that all we want are the critical points that are in the interval. Let’s do some examples. Example 1 Determine the absolute extrema for the following function and interval. ( ) [ ]3 22 3 12 4 on 4, 2g t t t t= + − + − Solution All we really need to do here is follow the procedure given above. So, first notice that this is a polynomial and so in continuous everywhere and in particular is then continuous on the given


Calculus I


interval. Now, we need to get the derivative so that we can find the critical points of the function. ( ) ( ) ( )26 6 12 6 2 1g t t t t t′ = + − = + − It looks like we’ll have two critical points, 2t = − and 1t = . Note that we actually want something more than just the critical points. We only want the critical points of the function that lie in the interval in question. Both of these do fall in the interval as so we will use both of them. That may seem like a silly thing to mention at this point, but it is often forgotten, usually when it becomes important, and so we will mention it at every opportunity to make sure it’s not forgotten. Now we evaluate the function at the critical points and the end points of the interval.

( ) ( )( ) ( )

2 24 1 3

4 28 2 8

g g

g g

− = = −

− = − =

Absolute extrema are the largest and smallest the function will ever be and these four points represent the only places in the interval where the absolute extrema can occur. So, from this list we see that the absolute maximum of g(t) is 24 and it occurs at 2t = − (a critical point) and the absolute minimum of g(t) is -28 which occurs at 4t = − (an endpoint). In this example we saw that absolute extrema can and will occur at both endpoints and critical points. One of the biggest mistakes that students make with these problems is to forget to check the endpoints of the interval. Example 2 Determine the absolute extrema for the following function and interval. ( ) [ ]3 22 3 12 4 on 0,2g t t t t= + − + Solution Note that this problem is almost identical to the first problem. The only difference is the interval that we’re working on. This small change will completely change our answer however. With this change we have excluded both of the answers from the first example. The first step is to again find the critical points. From the first example we know these are

2t = − and 1t = .. At this point it’s important to recall that we only want the critical points that actually fall in the interval in question. This means that we only want 1t = since 2t = − falls outside the interval. Now evaluate the function at the single critical point in the interval and the two endpoints. ( ) ( ) ( )1 3 0 4 2 8g g g= − = = From this list of values we see that the absolute maximum is 8 and will occur at 2t = and the absolute minimum is -3 which occurs at 1t = .


Calculus I


As we saw in this example a simple change in the interval can completely change the answer. It also has shown us that we do need to be careful to exclude critical points that aren’t in the interval. Had we forgotten this and included 2t = − we would have gotten the wrong absolute maximum! This is the other big mistakes that students make in these problems. All too often they forget to exclude critical points that aren’t in the interval. If your instructor is anything like me this will mean that you will get the wrong answer. It’s not too hard to make sure that a critical point outside of the interval is larger or smaller than any of the points in the interval. Example 3 Suppose that the population (in thousands) of a certain kind of insect after t months is given by the following formula. ( ) ( )3 sin 4 100P t t t= + + Determine the minimum and maximum population in the first 4 months. Solution The question that we’re really asking is to find the absolute extrema of P(t) on the interval [0,4]. Since this function is continuous everywhere we know we can do this. Let’s start with the derivative. ( ) ( )3 4cos 4P t t′ = + We need the critical points of the function. The derivative exists everywhere so there are no critical points from that. So, all we need to do is determine where the derivative is zero.

( )

( )

3 4cos 4 03cos 44

t

t

+ =

= −

The solutions to this are,

0.6047 , 0, 1, 2,4 2.4189 2 , 0, 1, 2, 2

4 3.8643 2 , 0, 1, 2, 0.9661 , 0, 1, 2,2

nt nt n nt n n nt n

πππ π

= + = ± ±= + = ± ±⇒

= + = ± ±= + = ± ±

……… …

So, these are all the critical points. We need to determine the ones that fall in the interval [0,4]. There’s nothing to do except plug some n’s into the formulas until we get all of them.

0n = : 0.6047 0.9661t t= = We’ll need both of these critical points.

1n = :


Calculus I


0.6047 2.1755 0.9661 2.53692 2

t tπ π= + = = + =

We’ll need these.

2n = : 0.6047 3.7463 0.9661 4.1077t tπ π= + = = + = In this case we only need the first one since the second is out of the interval. There are five critical points that are in the interval. They are, 0.6047, 0.9661, 2.1755, 2.5369, 3.7463 Finally, to determine the absolute minimum and maximum population we only need to plug these values into the function as well as the two end points. Here are the function evaluations.

( ) ( )( ) ( )( ) ( )( )

0 100.0 4 111.7121

0.6047 102.4756 0.9661 102.2368

2.1755 107.1880 2.5369 106.9492

3.7463 111.9004

P P

P P

P P

P

= =

= =

= =

=

From these evaluations it appears that the minimum population is 100,000 (remember that P is in thousands…) which occurs at 0t = and the maximum population is 111,900 which occurs at

3.7463t = . Make sure that you can correctly solve trig equations. If we had forgotten the 2 nπ we would have missed the last three critical points in the interval and hence gotten the wrong answer since the maximum population was at the final critical point. Also, note that we do really need to be very careful with rounding answers here. If we’d rounded to the nearest integer, for instance, it would appear that the maximum population would have occurred at two different locations instead of only one. Example 4 Suppose that the amount of money in a bank account after t years is given by,

( )2

582000 10t

A t t−

= − e Determine the minimum and maximum amount of money in the account during the first 10 years that it is open. Solution Here we are really asking for the absolute extrema of A(t) on the interval [0,10]. As with the previous examples this function is continuous everywhere and so we know that this can be done. We’ll first need the derivative so we can find the critical points.


Calculus I


( )

2 2

2

5 58 8

258

10 104

10 14

t t

t

tA t t

t

− −

−

′ = − − −

= − +

e e

e

The derivative exists everywhere and the exponential is never zero. Therefore the derivative will only be zero where,

2

21 0 4 24t t t− + = ⇒ = ⇒ = ±

We’ve got two critical points, however only 2t = is actually in the interval so that is only critical point that we’ll use. Let’s now evaluate the function at the lone critical point and the end points of the interval. Here are those function evaluations. ( ) ( ) ( )0 2000 2 199.66 10 1999.94A A A= = = So, the maximum amount in the account will be $2000 which occurs at 0t = and the minimum amount in the account will be $199.66 which occurs at the 2 year mark. In this example there are two important things to note. First, if we had included the second critical point we would have gotten an incorrect answer for the maximum amount so it’s important to be careful with which critical points to include and which to exclude. All of the problems that we’ve worked to this point had derivatives that existed everywhere and so the only critical points that we looked at were those for which the derivative is zero. Do not get too locked into this always happening. Most of the problems that we run into will be like this, but they won’t all be like this. Let’s work another example to make this point. Example 5 Determine the absolute extrema for the following function and interval.

( ) ( ) [ ]233 4 on 5, 1Q y y y= + − −

Solution Again, as with all the other examples here, this function is continuous on the given interval and so we know that this can be done. First we’ll need the derivative and make sure you can do the simplification that we did here to make the work for finding the critical points easier.


Calculus I


( ) ( ) ( )

( )( )

( )( )

( )

2 13 3

23

13

13

13

23 4 3 43

23 44

3 4 2

45 12

4

Q y y y y

yyy

y y

yy

y

− ′ = + + +

= + ++

+ +=

+

+=

+

So, it looks like we’ve got two critical points.

4 Because the derivative doesn't exist here.12 Because the derivative is zero here.5

y

y

= −

= −

Both of these are in the interval so let’s evaluate the function at these points and the end points of the interval.

( )

( ) ( )

124 0 9.8495

5 15 1 6.241

Q Q

Q Q

− = − = −

− = − − = −

The function has an absolute maximum of zero at 4y = − and the function will have an absolute minimum of -15 at 5y = − . So, if we had ignored or forgotten about the critical point where the derivative doesn’t exist ( 4y = − ) we would not have gotten the correct answer. In this section we’ve seen how we can use a derivative to identify the absolute extrema of a function. This is an important application of derivatives that will arise from time to time so don’t forget about it.


Calculus I


The Shape of a Graph, Part I In the previous section we saw how to use the derivative to determine the absolute minimum and maximum values of a function. However, there is a lot more information about a graph that can be determined from the first derivative of a function. We will start looking at that information in this section. The main idea we’ll be looking at in this section we will be identifying all the relative extrema of a function. Let’s start this section off by revisiting a familiar topic from the previous chapter. Let’s suppose that we have a function, ( )f x . We know from our work in the previous chapter that the first

derivative, ( )f x′ , is the rate of change of the function. We used this idea to identify where a

function was increasing, decreasing or not changing. Before reviewing this idea let’s first write down the mathematical definition of increasing and decreasing. We all know what the graph of an increasing/decreasing function looks like but sometimes it is nice to have a mathematical definition as well. Here it is. Definition 1. Given any 1x and 2x from an interval I with 1 2x x< if ( ) ( )1 2f x f x< then ( )f x is

increasing on I. 2. Given any 1x and 2x from an interval I with 1 2x x< if ( ) ( )1 2f x f x> then ( )f x is

decreasing on I. This definition will actually be used in the proof of the next fact in this section. Now, recall that in the previous chapter we constantly used the idea that if the derivative of a function was positive at a point then the function was increasing at that point and if the derivative was negative at a point then the function was decreasing at that point. We also used the fact that if the derivative of a function was zero at a point then the function was not changing at that point. We used these ideas to identify the intervals in which a function is increasing and decreasing. The following fact summarizes up what we were doing in the previous chapter. Fact 1. If ( ) 0f x′ > for every x on some interval I, then ( )f x is increasing on the interval.

2. If ( ) 0f x′ < for every x on some interval I, then ( )f x is decreasing on the interval.

3. If ( ) 0f x′ = for every x on some interval I, then ( )f x is constant on the interval. The proof of this fact is in the Proofs From Derivative Applications section of the Extras chapter.


Calculus I


Let’s take a look at an example. This example has two purposes. First, it will remind us of the increasing/decreasing type of problems that we were doing in the previous chapter. Secondly, and maybe more importantly, it will now incorporate critical points into the solution. We didn’t know about critical points in the previous chapter, but if you go back and look at those examples, the first step in almost every increasing/decreasing problem is to find the critical points of the function. Example 1 Determine all intervals where the following function is increasing or decreasing.

( ) 5 4 35 40 52 3

f x x x x= − + + +

Solution To determine if the function is increasing or decreasing we will need the derivative.

( )( )( )( )

4 3 2

2 2

2

5 10 40

5 2 8

5 4 2

f x x x x

x x x

x x x

′ = − + +

= − − −

= − − +

Note that when we factored the derivative we first factored a “-1” out to make the rest of the factoring a little easier. From the factored form of the derivative we see that we have three critical points : 2x = − ,

0x = , and 4x = . We’ll need these in a bit. We now need to determine where the derivative is positive and where it’s negative. We’ve done this several times now in both the Review chapter and the previous chapter. Since the derivative is a polynomial it is continuous and so we know that the only way for it to change signs is to first go through zero. In other words, the only place that the derivative may change signs is at the critical points of the function. We’ve now got another use for critical points. So, we’ll build a number line, graph the critical points and pick test points from each region to see if the derivative is positive or negative in each region. Here is the number line and the test points for the derivative.


Calculus I


Make sure that you test your points in the derivative. One of the more common mistakes here is to test the points in the function instead! Recall that we know that the derivative will be the same sign in each region. The only place that the derivative can change signs is at the critical points and we’ve marked the only critical points on the number line. So, it looks we’ve got the following intervals of increase and decrease.

Increase : 2 0 and 0 4Decrease : 2 and 4

x xx x

− < < < <− ∞ < < − < < ∞

In this example we used the fact that the only place that a derivative can change sign is at the critical points. Also, the critical points for this function were those for which the derivative was zero. However, the same thing can be said for critical points where the derivative doesn’t exist. This is nice to know. A function can change signs where it is zero or doesn’t exist. In the previous chapter all our examples of this type had only critical points where the derivative was zero. Now, that we know more about critical points we’ll also see an example or two later on with critical points where the derivative doesn’t exist. Now that we have the previous “reminder” example out of the way let’s move into some new material. Once we have the intervals of increasing and decreasing for a function we can use this information to get a sketch of the graph. Note that the sketch, at this point, may not be super accurate when it comes to the curvature of the graph, but it will at least have the basic shape correct. To get the curvature of the graph correct we’ll need the information from the next section. Let’s attempt to get a sketch of the graph of the function we used in the previous example. Example 2 Sketch the graph of the following function.

( ) 5 4 35 40 52 3

f x x x x= − + + +

Solution There really isn’t a whole lot to this example. Whenever we sketch a graph it’s nice to have a few points on the graph to give us a starting place. So we’ll start by the function at the critical points. These will give us some starting points when we go to sketch the graph. These points are,

( ) ( ) ( )89 14232 29.67 0 5 4 474.333 3

f f f− = − = − = = =

Once these points are graphed we go to the increasing and decreasing information and start sketching. For reference purposes here is the increasing/decreasing information.

Increase : 2 0 and 0 4Decrease : 2 and 4

x xx x

− < < < <− ∞ < < − < < ∞


Calculus I


Note that we are only after a sketch of the graph. As noted before we started this example we won’t be able to accurately predict the curvature of the graph at this point. However, even without this information we will still be able to get a basic idea of what the graph should look like. To get this sketch we start at the very left of the graph and knowing that the graph must be decreasing and will continue to decrease until we get to 2x = − . At this point the function will continue to increase until it gets to 4x = . However, note that during the increasing phase it does need to go through the point at 0x = and at this point we also know that the derivative is zero here and so the graph goes through 0x = horizontally. Finally, once we hit 4x = the graph starts, and continues, to decrease. Also, note that just like at 0x = the graph will need to be horizontal when it goes through the other two critical points as well. Here is the graph of the function. We, of course, used a graphical program to generate this graph, however, outside of some potential curvature issues if you followed the increasing/decreasing information and had all the critical points plotted first you should have something similar to this.

Let’s use the sketch from this example to give us a very nice test for classifying critical points as relative maximums, relative minimums or neither minimums or maximums. Recall Fermat’s Theorem from the Minimum and Maximum Values section. This theorem told us that all relative extrema (provided the derivative exists at that point of course) of a function will be critical points. The graph in the previous example has two relative extrema and both occur at critical points as the Fermat’s Theorem predicted. Note as well that we’ve got a critical point that isn’t a relative extrema ( 0x = ). This is okay since Fermat’s theorem doesn’t say that all critical points will be relative extrema. It only states that relative extrema will be critical points. In the sketch of the graph from the previous example we can see that to the left of 2x = − the graph is decreasing and to the right of 2x = − the graph is increasing and 2x = − is a relative minimum. In other words, the graph is behaving around the minimum exactly as it would have to


Calculus I


be in order for 2x = − to be a minimum. The same thing can be said for the relative maximum at 4x = . The graph is increasing on the left and decreasing on the right exactly as it must be in

order for 4x = to be a maximum. Finally, the graph is increasing on both sides of 0x = and so this critical point can’t be a minimum or a maximum. These ideas can be generalized to arrive at a nice way to test if a critical point is a relative minimum, relative maximum or neither. If x c= is a critical point and the function is decreasing to the left of x c= and is increasing to the right then x c= must be a relative minimum of the function. Likewise, if the function is increasing to the left of x c= and decreasing to the right then x c= must be a relative maximum of the function. Finally, if the function is increasing on both sides of x c= or decreasing on both sides of x c= then x c= can be neither a relative minimum nor a relative maximum. These ideas can be summarized up in the following test. First Derivative Test Suppose that x c= is a critical point of ( )f x then,

1. If ( ) 0f x′ > to the left of x c= and ( ) 0f x′ < to the right of x c= then x c= is a relative

maximum. 2. If ( ) 0f x′ < to the left of x c= and ( ) 0f x′ > to the right of x c= then x c= is a relative

minimum. 3. If ( )f x′ is the same sign on both sides of x c= then x c= is neither a relative maximum

nor a relative minimum. It is important to note here that the first derivative test will only classify critical points as relative extrema and not as absolute extrema. As we recall from the Finding Absolute Extrema section absolute extrema are largest and smallest function values and may not even exist or be critical points if they do exist. The first derivative test is exactly that, a test using the first derivative. It doesn’t ever use the value of the function and so no conclusions can be drawn from the test about the relative “size” of the function at the critical points (which would be needed to identify absolute extrema) and can’t even begin to address the fact that absolute extrema may not occur at critical points. Let’s take at another example. Example 3 Find and classify all the critical points of the following function. Give the intervals where the function is increasing and decreasing.

( ) 3 2 4g t t t= −


Calculus I


Solution First we’ll need the derivative so we can get our hands on the critical points. Note as well that we’ll do some simplification on the derivative to help us find the critical points.

( ) ( ) ( )

( )( )

( )( )

( )

1 22 2 23 3

1 22 3

22 3

2 2

22 3

2

22 3

24 43

243 4

3 4 2

3 4

5 12

3 4

g t t t t

ttt

t t

t

t

t

−′ = − + −

= − +−

− +=

−

−=

−

So, it looks like we’ll have four critical points here. They are,

2 The derivative doesn't exist here.

12 1.549 The derivative is zero here.5

t

t

= ±

= ± = ±

Finding the intervals of increasing and decreasing will also give the classification of the critical points so let’s get those first. Here is a number line with the critical points graphed and test points.

So, it looks like we’ve got the following intervals of increasing and decreasing.

12 12Increase : 2, 2 , 2, & 25 5

12 12Decrease : 5 5

x x x x

x

− ∞ < − − < < − < < < < ∞

− < <

From this it looks like 2t = − and 2t = are neither relative minimum or relative maximums

since the function is increasing on both side of them. On the other hand, 125t = − is a relative


Calculus I


maximum and 125t = is a relative minimum.

For completeness sake here is the graph of the function.

In the previous example the two critical points where the derivative didn’t exist ended up not being relative extrema. Do not read anything into this. They often will be relative extrema. Check out this example in the Absolute Extrema to see an example of one such critical point. Let’s work a couple more examples. Example 4 Suppose that the elevation above sea level of a road is given by the following function.

( ) 500 cos 3 sin4 4x xE x = + +

where x is in miles. Assume that if x is positive we are to the east of the initial point of measurement and if x is negative we are to the west of the initial point of measurement. If we start 25 miles to the west of the initial point of measurement and drive until we are 25 miles east of the initial point how many miles of our drive were we driving up an incline? Solution Okay, this is just a really fancy way of asking what the intervals of increasing and decreasing are for the function on the interval [-25,25]. So, we first need the derivative of the function.

( ) 1 3sin cos4 4 4 4

x xE x ′ = − +

Setting this equal to zero gives,


Calculus I


1 3sin cos 04 4 4 4

tan 34

x x

x

− + =

=

The solutions to this and hence the critical points are,

1.0472 2 , 0, 1, 2, 4.1888 8 , 0, 1, 2,4

16.7552 8 , 0, 1, 2,4.1888 2 , 0, 1, 2,4

x n n x n nx x n nn n

π ππ

π

= + = ± ± = + = ± ±⇒

= + = ± ±= + = ± ±

… ………

I’ll leave it to you to check that the critical points that fall in the interval that we’re after are, 20.9439, 8.3775, 4.1888, 16.7552− − Here is the number line with the critical points and test points.

So, it looks like the intervals of increasing and decreasing are,

Increase : 25 20.9439, 8.3775 4.1888 and 16.7552 25Decrease : 20.9439 8.3775 and 4.1888 16.7552

x x xx x

− < < − − < < < <− < < − < <

Notice that we had to end our intervals at -25 and 25 since we’ve done no work outside of these points and so we can’t really say anything about the function outside of the interval [-25,25]. From the intervals we can actually answer the question. We were driving on an incline during the intervals of increasing and so the total number of miles is,

( )( ) ( )( ) ( )Distance 20.9439 25 4.1888 8.3775 25 16.7552

24.8652 miles

= − − − + − − + −

=

Even though the problem didn’t ask for it we can also classify the critical points that are in the interval [-25,25].

Relative Maximums : 20.9439, 4.1888Relative Minimums : 8.3775,16.7552

−−


Calculus I


Example 5 The population of rabbits (in hundreds) after t years in a certain area is given by the following function, ( ) ( )2 ln 3 6P t t t= + Determine if the population ever decreases in the first two years. Solution So, again we are really after the intervals and increasing and decreasing in the interval [0,2]. We found the only critical point to this function back in the Critical Points section to be,

1 0.2023

x = =e

Here is a number line for the intervals of increasing and decreasing.

So, it looks like the population will decrease for a short period and then continue to increase forever. Also, while the problem didn’t ask for it we can see that the single critical point is a relative minimum. In this section we’ve seen how we can use the first derivative of a function to give us some information about the shape of a graph and how we can use this information in some applications. Using the first derivative to give us information about a whether a function is increasing or decreasing is a very important application of derivatives and arises on a fairly regular basis in many areas.


Calculus I


The Shape of a Graph, Part II In the previous section we saw how we could use the first derivative of a function to get some information about the graph of a function. In this section we are going to look at the information that the second derivative of a function can give us a about the graph of a function. Before we do this we will need a couple of definitions out of the way. The main concept that we’ll be discussing in this section is concavity. Concavity is easiest to see with a graph (we’ll give the mathematical definition in a bit).

So a function is concave up if it “opens” up and the function is concave down if it “opens” down. Notice as well that concavity has nothing to do with increasing or decreasing. A function can be concave up and either increasing or decreasing. Similarly, a function can be concave down and either increasing or decreasing. It’s probably not the best way to define concavity by saying which way it “opens” since this is a somewhat nebulous definition. Here is the mathematical definition of concavity. Definition 1 Given the function ( )f x then

1. ( )f x is concave up on an interval I if all of the tangents to the curve on I are below the

graph of ( )f x .

2. ( )f x is concave down on an interval I if all of the tangents to the curve on I are above

the graph of ( )f x . To show that the graphs above do in fact have concavity claimed above here is the graph again (blown up a little to make things clearer).


Calculus I


So, as you can see, in the two upper graphs all of the tangent lines sketched in are all below the graph of the function and these are concave up. In the lower two graphs all the tangent lines are above the graph of the function and these are concave down. Again, notice that concavity and the increasing/decreasing aspect of the function is completely separate and do not have anything to do with the other. This is important to note because students often mix these two up and use information about one to get information about the other. There’s one more definition that we need to get out of the way. Definition 2 A point x c= is called an inflection point if the function is continuous at the point and the concavity of the graph changes at that point. Now that we have all the concavity definitions out of the way we need to bring the second derivative into the mix. We did after all start off this section saying we were going to be using the second derivative to get information about the graph. The following fact relates the second derivative of a function to its concavity. The proof of this fact is in the Proofs From Derivative Applications section of the Extras chapter. Fact Given the function ( )f x then,

1. If ( ) 0f x′′ > for all x in some interval I then ( )f x is concave up on I.

2. If ( ) 0f x′′ < for all x in some interval I then ( )f x is concave down on I.


Calculus I


Notice that this fact tells us that a list of possible inflection points will be those points where the second derivative is zero or doesn’t exist. Be careful however to not make the assumption that just because the second derivative is zero or doesn’t exist that the point will be an inflection point. We will only know that it is an inflection point once we determine the concavity on both sides of it. It will only be an inflection point if the concavity is different on both sides of the point. Now that we know about concavity we can use this information as well as the increasing/decreasing information from the previous section to get a pretty good idea of what a graph should look like. Let’s take a look at an example of that. Example 1 For the following function identify the intervals where the function is increasing and decreasing and the intervals where the function is concave up and concave down. Use this information to sketch the graph. ( ) 5 33 5 3h x x x= − + Solution Okay, we are going to need the first two derivatives so let’s get those first.

( ) ( ) ( )( ) ( )

4 2 2

3 2

15 15 15 1 1

60 30 30 2 1

h x x x x x x

h x x x x x

′ = − = − +

′′ = − = −

Let’s start with the increasing/decreasing information since we should be fairly comfortable with that after the last section. There are three critical points for this function : 1x = − , 0x = , and 1x = . Below is the number line for the increasing/decreasing information.

So, it looks like we’ve got the following intervals of increasing and decreasing.

Increasing : 1 and 1Decreasing : 1 0, 0 1

x xx x

− ∞ < < − < < ∞− < < < <

Note that from the first derivative test we can also say that 1x = − is a relative maximum and that 1x = is a relative minimum. Also 0x = is neither a relative minimum or maximum. Now let’s get the intervals where the function is concave up and concave down. If you think about it this process is almost identical to the process we use to identify the intervals of increasing


Calculus I


and decreasing. This only difference is that we will be using the second derivative instead of the first derivative. The first thing that we need to do is identify the possible inflection points. These will be where the second derivative is zero or doesn’t exist. The second derivative in this case is a polynomial and so will exist everywhere. It will be zero at the following points.

10, 0.70712

x x= = ± = ±

As with the increasing and decreasing part we can draw a number line up and use these points to divide the number line into regions. In these regions we know that the second derivative will always have the same sign since these three points are the only places where the function may change sign. Therefore, all that we need to do is pick a point from each region and plug it into the second derivative. The second derivative will then have that sign in the whole region from which the point came from Here is the number line for this second derivative.

So, it looks like we’ve got the following intervals of concavity.

1 1Concave Up : 0 and 2 2

1 1Concave Down : and 02 2

x x

x x

− < < < < ∞

− ∞ < < − < <

This also means that

10, 0.70712

x x= = ± = ±

are all inflection points. All this information can be a little overwhelming when going to sketch the graph. The first thing that we should do is get some starting points. The critical points and inflection points are good starting points. So, first graph these points. Now, start to the left and start graphing the increasing/decreasing information as we did in the previous section when all we had was the increasing/decreasing information. As we graph this we will make sure that the concavity information matches up with what we’re graphing.


Calculus I


Using all this information to sketch the graph gives the following graph.

We can use the previous example to illustrate another way to classify some of the critical points of a function as relative maximums or relative minimums. Notice that 1x = − is a relative maximum and that the function is concave down at this point. This means that ( )1f ′′ − must be negative. Likewise, 1x = is a relative minimum and the

function is concave up at this point. This means that ( )1f ′′ must be positive.

As we’ll see in a bit we will need to be very careful with 0x = . In this case the second derivative is zero, but that will not actually mean that 0x = is not a relative minimum or maximum. We’ll see some examples of this in a bit, but we need to get some other information taken care of first. It is also important to note here that all of the critical points in this example were critical points in which the first derivative was zero and this is required for this to work. We will not be able to use this test on critical points where the derivative doesn’t exist. Here is the test that can be used to classify some of the critical points of a function. The proof of this test is in the Proofs From Derivative Applications section of the Extras chapter. Second Derivative Test Suppose that x c= is a critical point of ( )f c′ such that ( ) 0f c′ = and that ( )f x′′ is

continuous in a region around x c= . Then, 1. If ( ) 0f c′′ < then x c= is a relative maximum.

2. If ( ) 0f c′′ > then x c= is a relative minimum.

3. If ( ) 0f c′′ = then x c= can be a relative maximum, relative minimum or neither.


Calculus I


The third part of the second derivative test is important to notice. If the second derivative is zero then the critical point can be anything. Below are the graphs of three functions all of which have a critical point at 0x = , the second derivative of all of the functions is zero at 0x = and yet all three possibilities are exhibited. The first is the graph of ( ) 4f x x= . This graph has a relative minimum at 0x = .

Next is the graph of ( ) 4f x x= − which has a relative maximum at 0x = .

Finally, there is the graph of ( ) 3f x x= and this graph had neither a relative minimum or a

relative maximum at 0x = .


Calculus I


So, we can see that we have to be careful if we fall into the third case. For those times when we do fall into this case we will have to resort to other methods of classifying the critical point. This is usually done with the first derivative test. Let’s go back and relook at the critical points from the first example and use the Second Derivative Test on them, if possible. Example 2 Use the second derivative test to classify the critical points of the function, ( ) 5 33 5 3h x x x= − + Solution Note that all we’re doing here is verifying the results from the first example. The second derivative is, ( ) 360 30h x x x′′ = − The three critical points ( 1x = − , 0x = , and 1x = ) of this function are all critical points where the first derivative is zero so we know that we at least have a chance that the Second Derivative Test will work. The value of the second derivative for each of these are, ( ) ( ) ( )1 30 0 0 1 30h h h′′ ′′ ′′− = − = = The second derivative at 1x = − is negative so by the Second Derivative Test this critical point this is a relative maximum as we saw in the first example. The second derivative at 1x = is positive and so we have a relative minimum here by the Second Derivative Test as we also saw in the first example. In the case of 0x = the second derivative is zero and so we can’t use the Second Derivative Test to classify this critical point. Note however, that we do know from the First Derivative Test we used in the first example that in this case the critical point is not a relative extrema.


Calculus I


Let’s work one more example. Example 3 For the following function find the inflection points and use the second derivative test, if possible, to classify the critical points. Also, determine the intervals of increase/decrease and the intervals of concave up/concave down and sketch the graph of the function.

( ) ( )236f t t t= −

Solution We’ll need the first and second derivatives to get us started.

( )( )

( )( )

1 43 3

18 5 10 72

3 6 9 6

t tf t f tt t

− −′ ′′= =− −

The critical points are,

18 3.6 65

t t= = =

Notice as well that we won’t be able to use the second derivative test on 6t = to classify this critical point since the derivative doesn’t exist at this point. To classify this we’ll need the increasing/decreasing information that we’ll get to sketch the graph. We can however, use the Second Derivative Test to classify the other critical point so let’s do that before we proceed with the sketching work. Here is the value of the second derivative at 3.6t = . ( )3.6 1.245 0f ′′ = − < So, according to the second derivative test 3.6t = is a relative maximum. Now let’s proceed with the work to get the sketch of the graph and notice that once we have the increasing/decreasing information we’ll be able to classify 6t = . Here is the number line for the first derivative.

So, according to the first derivative test we can verify that 3.6t = is in fact a relative maximum. We can also see that 6t = is a relative minimum. Be careful not to assume that a critical point that can’t be used in the second derivative test won’t be a relative extrema. We’ve clearly seen now both with this example and in the discussion after


Calculus I


we have the test that just because we can’t use the Second Derivative Test or the Test doesn’t tell us anything about a critical point doesn’t mean that the critical point will not be a relative extrema. This is a common mistake that many students make so be careful when using the Second Derivative Test. Okay, let’s finish the problem out. We will need the list of possible inflection points. These are,

726 7.210

t t= = =

Here is the number line for the second derivative. Note that we will need this to see if the two points above are in fact inflection points.

So, the concavity only changes at 7.2t = and so this is the only inflection point for this function. Here is the sketch of the graph.

The change of concavity at 7.2t = is hard to see, but it is there it’s just a very subtle change in concavity.


Calculus I


The Mean Value Theorem In this section we want to take a look at the Mean Value Theorem. In most traditional textbooks this section comes before the sections containing the First and Second Derivative Tests because many of the proofs in those sections need the Mean Value Theorem. However, we feel that from a logical point of view it’s better to put the Shape of a Graph sections right after the absolute extrema section. So, if you’ve been following the proofs from the previous two sections you’ve probably already read through this section. Before we get to the Mean Value Theorem we need to cover the following theorem. Rolle’s Theorem Suppose ( )f x is a function that satisfies all of the following.

1. ( )f x is continuous on the closed interval [a,b].

2. ( )f x is differentiable on the open interval (a,b).

3. ( ) ( )f a f b=

Then there is a number c such that a c b< < and ( ) 0f c′ = . Or, in other words ( )f x has a

critical point in (a,b). To see the proof of Rolle’s Theorem see the Proofs From Derivative Applications section of the Extras chapter. Let’s take a look at a quick example that uses Rolle’s Theorem. Example 1 Show that ( ) 5 34 7 2f x x x x= + + − has exactly one real root. Solution From basic Algebra principles we know that since ( )f x is a 5th degree polynomial it will have

five roots. What we’re being asked to prove here is that only one of those 5 is a real number and the other 4 must be complex roots. First, we should show that it does have at least one real root. To do this note that ( )0 2f = − and

that ( )1 10f = and so we can see that ( ) ( )0 0 1f f< < . Now, because ( )f x is a polynomial

we know that it is continuous everywhere and so by the Intermediate Value Theorem there is a number c such that 0 1c< < and ( ) 0f c = . In other words ( )f x has at least one real root.

We now need to show that this is in fact the only real root. To do this we’ll use an argument that


Calculus I


is called contradiction proof. What we’ll do is assume that ( )f x has at least two real roots.

This means that we can find real numbers a and b (there might be more, but all we need for this particular argument is two) such that ( ) ( ) 0f a f b= = . But if we do this then we know from

Rolle’s Theorem that there must then be another number c such that ( ) 0f c′ = .

This is a problem however. The derivative of this function is, ( ) 4 220 3 7f x x x′ = + +

Because the exponents on the first two terms are even we know that the first two terms will always be greater than or equal to zero and we are then going to add a positive number onto that and so we can see that the smallest the derivative will ever be is 7 and this contradicts the statement above that says we MUST have a number c such that ( ) 0f c′ = .

We reached these contradictory statements by assuming that ( )f x has at least two roots. Since

this assumption leads to a contradiction the assumption must be false and so we can only have a single real root. The reason for covering Rolle’s Theorem is that it is needed in the proof of the Mean Value Theorem. To see the proof see the Proofs From Derivative Applications section of the Extras chapter. Here is the theorem. Mean Value Theorem Suppose ( )f x is a function that satisfies both of the following.

1. ( )f x is continuous on the closed interval [a,b].

2. ( )f x is differentiable on the open interval (a,b). Then there is a number c such that a < c < b and

( ) ( ) ( )f b f af c

b a−

′ =−

Or, ( ) ( ) ( )( )f b f a f c b a′− = − Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c that will satisfy the conclusion of the theorem. Also note that if it weren’t for the fact that we needed Rolle’s Theorem to prove this we could think of Rolle’s Theorem as a special case of the Mean Value Theorem. To see that just assume that ( ) ( )f a f b= and then the result of the Mean Value Theorem gives the result of Rolle’s

Theorem.


Calculus I


Before we take a look at a couple of examples let’s think about a geometric interpretation of the Mean Value Theorem. First define ( )( ),A a f a= and ( )( ),B b f b= and then we know from

the Mean Value theorem that there is a c such that a c b< < and that

( ) ( ) ( )f b f af c

b a−

′ =−

Now, if we draw in the secant line connecting A and B then we can know that the slope of the secant line is,

( ) ( )f b f ab a

−−

Likewise, if we draw in the tangent line to ( )f x at x c= we know that its slope is ( )f c′ .

What the Mean Value Theorem tells us is that these two slopes must be equal or in other words the secant line connecting A and B and the tangent line at x c= must be parallel. We can see this in the following sketch.

Let’s now take a look at a couple of examples using the Mean Value Theorem. Example 2 Determine all the numbers c which satisfy the conclusions of the Mean Value Theorem for the following function. ( ) [ ]3 22 on 1,2f x x x x= + − − Solution There isn’t really a whole lot to this problem other than to notice that since ( )f x is a polynomial

it is both continuous and differentiable (i.e. the derivative exists) on the interval given. First let’s find the derivative.


Calculus I


( ) 23 4 1f x x x′ = + − Now, to find the numbers that satisfy the conclusions of the Mean Value Theorem all we need to do is plug this into the formula given by the Mean Value Theorem.

( ) ( ) ( )

( )2

2 12 1

14 2 123 4 1 43 3

f ff c

c c

− −′ =

− −

−+ − = = =

Now, this is just a quadratic equation,

2

2

3 4 1 43 4 5 0c cc c

+ − =

+ − =

Using the quadratic formula on this we get,

( )( )4 16 4 3 5 4 76

6 6c

− ± − − − ±= =

So, solving gives two values of c.

4 76 4 760.7863 2.11966 6

c c− + − −= = = = −

Notice that only one of these is actually in the interval given in the problem. That means that we will exclude the second one (since it isn’t in the interval). The number that we’re after in this problem is, 0.7863c = Be careful to not assume that only one of the numbers will work. It is possible for both of them to work. Example 3 Suppose that we know that ( )f x is continuous and differentiable on [6, 15]. Let’s

also suppose that we know that ( )6 2f = − and that we know that ( ) 10f x′ ≤ . What is the

largest possible value for ( )15f ? Solution Let’s start with the conclusion of the Mean Value Theorem. ( ) ( ) ( )( )15 6 15 6f f f c′− = − Plugging in for the known quantities and rewriting this a little gives, ( ) ( ) ( ) ( ) ( )15 6 15 6 2 9f f f c f c′ ′= + − = − +


Calculus I


Now we know that ( ) 10f x′ ≤ so in particular we know that ( ) 10f c′ ≤ . This gives us the

following,

( ) ( )

( )15 2 9

2 9 1088

f f c′= − +

≤ − +

=

All we did was replace ( )f c′ with its largest possible value.

This means that the largest possible value for ( )15f is 88.

Example 4 Suppose that we know that ( )f x is continuous and differentiable everywhere.

Let’s also suppose that we know that ( )f x has two roots. Show that ( )f x′ must have at least one root. Solution It is important to note here that all we can say is that ( )f x′ will have at least one root. We can’t

say that it will have exactly one root. So don’t confuse this problem with the first one we worked. This is actually a fairly simple thing to prove. Since we know that ( )f x has two roots let’s

suppose that they are a and b. Now, by assumption we know that ( )f x is continuous and

differentiable everywhere and so in particular it is continuous on [a,b] and differentiable on (a,b). Therefore, by the Mean Value Theorem there is a number c that is between a and b (this isn’t needed for this problem, but it’s true so it should be pointed out) and that,

( ) ( ) ( )f b f af c

b a−

′ =−

But we now need to recall that a and b are roots of ( )f x and so this is,

( ) 0 0 0f cb a

−′ = =−

Or, ( )f x′ has a root at x c= .

Again, it is important to note that we don’t have a value of c. We have only shown that it exists. We also haven’t said anything about c being the only root. It is completely possible for ( )f x′ to

have more than one root.


Calculus I


It is completely possible to generalize the previous example significantly. For instance if we know that ( )f x is continuous and differentiable everywhere and has three roots we can then

show that not only will ( )f x′ have at least two roots but that ( )f x′′ will have at least one root.

We’ll leave it to you to verify this, but the ideas involved are identical to those in the previous example. We’ll close this section out with a couple of nice facts that can be proved using the Mean Value Theorem. Note that in both of these facts we are assuming the functions are continuous and differentiable on the interval [a,b]. Fact 1 If ( ) 0f x′ = for all x in an interval ( ),a b then ( )f x is constant on ( ),a b .

This fact is very easy to prove so let’s do that here. Take any two x’s in the interval ( ),a b , say

1x and 2x . Then since ( )f x is continuous and differential on [a,b] it must also be continuous

and differentiable on [ ]1 2,x x . This means that we can apply the Mean Value Theorem for these

two values of x. Doing this gives, ( ) ( ) ( )( )2 1 2 1f x f x f c x x′− = −

where 1 2x c x< < . But by assumption ( ) 0f x′ = for all x in an interval ( ),a b and so in

particular we must have, ( ) 0f c′ = Putting this into the equation above gives, ( ) ( ) ( ) ( )2 1 2 10f x f x f x f x− = ⇒ = Now, since 1x and 2x where any two values of x in the interval ( ),a b we can see that we must

have ( ) ( )2 1f x f x= for all 1x and 2x in the interval and this is exactly what it means for a

function to be constant on the interval and so we’ve proven the fact. Fact 2 If ( ) ( )f x g x′ ′= for all x in an interval ( ),a b then in this interval we have ( ) ( )f x g x c= +

where c is some constant. This fact is a direct result of the previous fact and is also easy to prove. If we first define,


Calculus I


( ) ( ) ( )h x f x g x= −

Then since both ( )f x and ( )g x are continuous and differentiable in the interval ( ),a b then so

must be ( )h x . Therefore the derivative of ( )h x is,

( ) ( ) ( )h x f x g x′ ′ ′= − However, by assumption ( ) ( )f x g x′ ′= for all x in an interval ( ),a b and so we must have that

( ) 0h x′ = for all x in an interval ( ),a b . So, by Fact 1 ( )h x must be constant on the interval.

This means that we have,

( )

( ) ( )( ) ( )

h x c

f x g x c

f x g x c

=

− =

= +

which is what we were trying to show.


Calculus I


Optimization In this section we are going to look at optimization problems. In optimization problems we are looking for the largest value or the smallest value that a function can take. We saw how to solve one kind of optimization problem in the Absolute Extrema section where we found the largest and smallest value that a function would take on an interval. In this section we are going to look at another type of optimization problem. Here we will be looking for the largest or smallest value of a function subject to some kind of constraint. The constraint will be some condition (that can usually be described by some equation) that must absolutely, positively be true no matter what our solution is. On occasion, the constraint will not be easily described by an equation, but in these problems it will be easy to deal with as we’ll see. This section is generally one of the more difficult for students taking a Calculus course. One of the main reasons for this is that a subtle change of wording can completely change the problem. There is also the problem of identifying the quantity that we’ll be optimizing and the quantity that is the constraint and writing down equations for each. The first step in all of these problems should be to very carefully read the problem. Once you’ve done that the next step is to identify the quantity to be optimized and the constraint. In identifying the constraint remember that the constraint is the quantity that must true regardless of the solution. In almost every one of the problems we’ll be looking at here one quantity will be clearly indicated as having a fixed value and so must be the constraint. Once you’ve got that identified the quantity to be optimized should be fairly simple to get. It is however easy to confuse the two if you just skim the problem so make sure you carefully read the problem first! Let’s start the section off with a simple problem to illustrate the kinds of issues we will be dealing with here. Example 1 We need to enclose a field with a fence. We have 500 feet of fencing material and a building is on one side of the field and so won’t need any fencing. Determine the dimensions of the field that will enclose the largest area. Solution In all of these problems we will have two functions. The first is the function that we are actually trying to optimize and the second will be the constraint. Sketching the situation will often help us to arrive at these equations so let’s do that.

In this problem we want to maximize the area of a field and we know that will use 500 ft of


Calculus I


fencing material. So, the area will be the function we are trying to optimize and the amount of fencing is the constraint. The two equations for these are,

Maximize : Contraint : 500 2

A xyx y

== +

Okay, we know how to find the largest or smallest value of a function provided it’s only got a single variable. The area function (as well as the constraint) has two variables in it and so what we know about finding absolute extrema won’t work. However, if we solve the constraint for one of the two variables we can substitute this into the area and we will then have a function of a single variable. So, let’s solve the constraint for x. Note that we could have just as easily solved for y but that would have led to fractions and so, in this case, solving for x will probably be best. 500 2x y= − Substituting this into the area function gives a function of y. ( ) ( ) 2500 2 500 2A y y y y y= − = − Now we want to find the largest value this will have on the interval [0,250]. Note that the interval corresponds to taking 0y = (i.e. no sides to the fence) and 250y = (i.e. only two sides and no width, also if there are two sides each must be 250 ft to use the whole 500ft). Note that the endpoints of the interval won’t make any sense from a physical standpoint if we actually want to enclose some area because they would both give zero area. They do, however, give us a set of limits on y and so the Extreme Value Theorem tells us that we will have a maximum value of the area somewhere between the two endpoints. Having these limits will also mean that we can use the process we discussed in the Finding Absolute Extrema section earlier in the chapter to find the maximum value of the area. So, recall that the maximum value of a continuous function (which we’ve got here) on a closed interval (which we also have here) will occur at critical points and/or end points. As we’ve already pointed out the end points in this case will give zero area and so don’t make any sense. That means our only option will be the critical points. So let’s get the derivative and find the critical points. ( ) 500 4A y y′ = − Setting this equal to zero and solving gives a lone critical point of 125y = . Plugging this into the area gives an area of 31250 ft2. So according to the method from Absolute Extrema section this must be the largest possible area, since the area at either endpoint is zero.


Calculus I


Finally, let’s not forget to get the value of x and then we’ll have the dimensions since this is what the problem statement asked for. We can get the x by plugging in our y into the constraint. ( )500 2 125 250x = − = The dimensions of the field that will give the largest area, subject to the fact that we used exactly 500 ft of fencing material, are 250 x 125. Don’t forget to actually read the problem and give the answer that was asked for. These types of problems can take a fair amount of time/effort to solve and it’s not hard to sometimes forget what the problem was actually asking for. In the previous problem we used the method from the Finding Absolute Extrema section to find the maximum value of the function we wanted to optimize. However, as we’ll see in later examples it will not always be easy to find endpoints. Also, even if we can find the endpoints we will see that sometimes dealing with the endpoints may not be easy either. Not only that, but this method requires that the function we’re optimizing be continuous on the interval we’re looking at, including the endpoints, and that may not always be the case. So, before proceeding with any more examples let’s spend a little time discussing some methods for determining if our solution is in fact the absolute minimum/maximum value that we’re looking for. In some examples all of these will work while in others one or more won’t be all that useful. However, we will always need to use some method for making sure that our answer is in fact that optimal value that we’re after. Method 1 : Use the method used in Finding Absolute Extrema. This is the method used in the first example above. Recall that in order to use this method the range of possible optimal values, let’s call it I, must have finite endpoints. Also, the function we’re optimizing (once it’s down to a single variable) must be continuous on I, including the endpoints. If these conditions are met then we know that the optimal value, either the maximum or minimum depending on the problem, will occur at either the endpoints of the range or at a critical point that is inside the range of possible solutions. There are two main issues that will often prevent this method from being used however. First, not every problem will actually have a range of possible solutions that have finite endpoints at both ends. We’ll see at least one example of this as we work through the remaining examples. Also, many of the functions we’ll be optimizing will not be continuous once we reduce them down to a single variable and this will prevent us from using this method. Method 2 : Use a variant of the First Derivative Test. In this method we also will need a range of possible optimal values, I. However, in this case, unlike the previous method the endpoints do not need to be finite. Also, we will need to require that the function be continuous on the interior I and we will only need the function to be


Calculus I


continuous at the end points if the endpoint is finite and the function actually exists at the endpoint. We’ll see several problems where the function we’re optimizing doesn’t actually exist at one of the endpoints. This will not prevent this method from being used. Let’s suppose that x c= is a critical point of the function we’re trying to optimize, ( )f x . We

already know from the First Derivative Test that if ( ) 0f x′ > immediately to the left of x c=

(i.e. the function is increasing immediately to the left) and if ( ) 0f x′ < immediately to the right

of x c= (i.e. the function is decreasing immediately to the right) then x c= will be a relative maximum for ( )f x .

Now, this does not mean that the absolute maximum of ( )f x will occur at x c= . However,

suppose that we knew a little bit more information. Suppose that in fact we knew that ( ) 0f x′ >

for all x in I such that x c< . Likewise, suppose that we knew that ( ) 0f x′ < for all x in I such

that x c> . In this case we know that to the left of x c= , provided we stay in I of course, the function is always increasing and to the right of x c= , again staying in I, we are always decreasing. In this case we can say that the absolute maximum of ( )f x in I will occur at x c= .

Similarly, if we know that to the left of x c= the function is always decreasing and to the right of x c= the function is always increasing then the absolute minimum of ( )f x in I will occur at

x c= . Before we give a summary of this method let’s discuss the continuity requirement a little. Nowhere in the above discussion did the continuity requirement apparently come into play. We require that the function we’re optimizing to be continuous in I to prevent the following situation.


Calculus I


In this case, a relative maximum of the function clearly occurs at x c= . Also, the function is always decreasing to the right and is always increasing to the left. However, because of the discontinuity at x d= , we can clearly see that ( ) ( )f d f c> and so the absolute maximum of

the function does not occur at x c= . Had the discontinuity at x d= not been there this would not have happened and the absolute maximum would have occurred at x c= . Here is a summary of this method. First Derivative Test for Absolute Extrema Let I be the interval of all possible optimal values of ( )f x and further suppose that ( )f x is

continuous on I , except possibly at the endpoints. Finally suppose that x c= is a critical point of ( )f x and that c is in the interval I. If we restrict x to values from I (i.e. we only consider

possible optimal values of the function) then,

1. If ( ) 0f x′ > for all x c< and if ( ) 0f x′ < for all x c> then ( )f c will be the

absolute maximum value of ( )f x on the interval I.

2. If ( ) 0f x′ < for all x c< and if ( ) 0f x′ > for all x c> then ( )f c will be the

absolute minimum value of ( )f x on the interval I. Method 3 : Use the second derivative. There are actually two ways to use the second derivative to help us identify the optimal value of a function and both use the Second Derivative Test to one extent or another. The first way to use the second derivative doesn’t actually help us to identify the optimal value. What it does do is allow us to potentially exclude values and knowing this can simplify our work somewhat and so is not a bad thing to do. Suppose that we are looking for the absolute maximum of a function and after finding the critical points we find that we have multiple critical points. Let’s also suppose that we run all of them through the second derivative test and determine that some of them are in fact relative minimums of the function. Since we are after the absolute maximum we know that a maximum (of any kind) can’t occur at relative minimums and so we immediately know that we can exclude these points from further consideration. We could do a similar check if we were looking for the absolute minimum. Doing this may not seem like all that great of a thing to do, but it can, on occasion, lead to a nice reduction in the amount of work that we need to do in later steps. The second way of using the second derivative to identify the optimal value of a function is in fact very similar to the second method above. In fact we will have the same requirements for this method as we did in that method. We need an interval of possible optimal values, I and the


Calculus I


endpoint(s) may or may not be finite. We’ll also need to require that the function, ( )f x be

continuous everywhere in I except possibly at the endpoints as above. Now, suppose that x c= is a critical point and that ( ) 0f c′′ > . The second derivative test tells

us that x c= must be a relative minimum of the function. Suppose however that we also knew that ( ) 0f x′′ > for all x in I. In this case we would know that the function was concave up in all

of I and that would in turn mean that the absolute minimum of ( )f x in I would in fact have to

be at x c= . Likewise if x c= is a critical point and ( ) 0f x′′ < for all x in I then we would know that the

function was concave down in I and that the absolute maximum of ( )f x in I would have to be

at x c= . Here is a summary of this method. Second Derivative Test for Absolute Extrema Let I be the range of all possible optimal values of ( )f x and further suppose that ( )f x is

continuous on I , except possibly at the endpoints. Finally suppose that x c= is a critical point of ( )f x and that c is in the interval I. Then,

1. If ( ) 0f x′′ > for all x in I then ( )f c will be the absolute minimum value of ( )f x on

the interval I. 2. If ( ) 0f x′′ < for all x in I then ( )f c will be the absolute maximum value of ( )f x on

the interval I. Before proceeding with some more examples we need to once again acknowledge that not every method discussed above will work for every problem and that, in some problems, more than one method will work. There are also problems where we may need to use a combination of these methods to identify the optimal value. Each problem will be different and we’ll need to see what we’ve got once we get the critical points before we decide which method might be best to use. Okay, let’s work some more examples. Example 2 We want to construct a box whose base length is 3 times the base width. The material used to build the top and bottom cost $10/ft2 and the material used to build the sides cost $6/ft2. If the box must have a volume of 50ft3 determine the dimensions that will minimize the cost to build the box. Solution First, a quick figure (probably not to scale…).


Calculus I


We want to minimize the cost of the materials subject to the constraint that the volume must be 50ft3. Note as well that the cost for each side is just the area of that side times the appropriate cost. The two functions we’ll be working with here this time are,

( ) ( ) 2

2

Minimize : 10 2 6 2 2 60 48

Constraint : 50 3

C lw wh lh w wh

lwh w h

= + + = +

= =

As with the first example, we will solve the constraint for one of the variables and plug this into the cost. It will definitely be easier to solve the constraint for h so let’s do that.

2

503

hw

=

Plugging this into the cost gives,

( ) 2 22

50 80060 48 603

C w w w ww w

= + = +

Now, let’s get the first and second (we’ll be needing this later…) derivatives,

( ) ( )3

2 32

120 800120 800 120 1600wC w w w C w ww

− −−′ ′′= − = = +

So, it looks like we’ve got two critical points here. The first is obvious, 0w = , and it’s also just as obvious that this will not be the correct value. We are building a box here and w is the box’s width and so since it makes no sense to talk about a box with zero width we will ignore this critical point. This does not mean however that you should just get into the habit of ignoring zero when it occurs. There are other types of problems where it will be a valid point that we will need to consider. The next critical point will come from determining where the numerator is zero.

3 3 3800 20120 800 0 1.8821120 3

w w− = ⇒ = = =


Calculus I


So, once we throw out 0w = , we’ve got a single critical point and we now have to verify that this is in fact the value that will give the absolute minimum cost. In this case we can’t use Method 1 from above. First, the function is not continuous at one of the endpoints, 0w = , of our interval of possible values. Secondly, there is no theoretical upper limit to the width that will give a box with volume of 50 ft3. If w is very large then we would just need to make h very small. The second method listed above would work here, but that’s going to involve some calculations, not difficult calculations, but more work nonetheless. The third method however, will work quickly and simply here. First, we know that whatever the value of w that we get it will have to be positive and we can see second derivative above that provided 0w > we will have ( ) 0C w′′ > and so in the interval of possible optimal values the

cost function will always be concave up and so 1.8821w = must give the absolute minimum cost. All we need to do now is to find the remaining dimensions.

( )

( )22

1.88213 3 1.8821 5.646350 50 4.7050

3 3 1.8821

wl w

hw

=

= = =

= = =

Also, even though it was not asked for, the minimum cost is : ( )1.8821 $637.60C = .

Example 3 We want to construct a box with a square base and we only have 10 m2 of material to use in construction of the box. Assuming that all the material is used in the construction process determine the maximum volume that the box can have. Solution This example is in many ways the exact opposite of the previous example. In this case we want to optimize the volume and the constraint this time is the amount of material used. We don’t have a cost here, but if you think about it the cost is nothing more than the amount of material used times a cost and so the amount of material and cost are pretty much tied together. If you can do one you can do the other as well. Note as well that the amount of material used is really just the surface area of the box. As always, let’s start off with a quick sketch of the box.


Calculus I


Now, as mentioned above we want to maximize the volume and the amount of material is the constraint so here are the equations we’ll need.

2

2

Maximize : Constraint : 10 2 2 2 2 4

V lwh w hlw wh lh w wh

= =

= + + = +

We’ll solve the constraint for h and plug this into the equation for the volume.

( ) ( )2 2 2

2 310 2 5 5 1 54 2 2 2

w w wh V w w w ww w w

− − −= = ⇒ = = −

Here are the first and second derivatives of the volume function.

( ) ( ) ( )212 5 3 3V w w V w w′ ′′= − = −

Note as well here that provided 0w > , which we know from a physical standpoint will be true, then the volume function will be concave down and so if we get a single critical point then we know that it will have to be the value that gives the absolute maximum. Setting the first derivative equal to zero and solving gives us the two critical points,

5 1.29103

w = ± = ±

In this case we can exclude the negative critical point since we are dealing with a length of a box and we know that these must be positive. Do not however get into the habit of just excluding any negative critical point. There are problems where negative critical points are perfectly valid possible solutions. Now, as noted above we got a single critical point, 1.2910, and so this must be the value that gives the maximum volume and since the maximum volume is all that was asked for in the problem statement the answer is then : ( ) 31.2910 2.1517 mV = .

Note that we could also have noted here that if 1.2910w < then ( ) 0V w′ > and likewise if


Calculus I


1.2910w > then ( ) 0V w′ < and so if we are to the left of the critical point the volume is always

increasing and if we are to the right of the critical point the volume is always decreasing and so by the Method 2 above we can also see that the single critical point must give the absolute maximum of the volume. Finally, even though these weren’t asked for here are the dimension of the box that gives the maximum volume.

( )25 1.29101.2910 1.2910

2 1.2910l w h −

= = = =

So, it looks like in this case we actually have a perfect cube. In the last two examples we’ve seen that many of these optimization problems can be done in both directions so to speak. In both examples we have essentially the same two equations: volume and surface area. However, in Example 2 the volume was the constraint and the cost (which is directly related to the surface area) was the function we were trying to optimize. In Example 3, on the other hand, we were trying to optimize the volume and the surface area was the constraint. It is important to not get so locked into one way of doing these problems that we can’t do it in the opposite direction as needed as well. This is one of the more common mistakes that students make with these kinds of problems. They see one problem and then try to make every other problem that seems to be the same conform to that one solution even if the problem needs to be worked differently. Keep an open mind with these problems and make sure that you understand what is being optimized and what the constraint is before you jump into the solution. Also, as seen in the last example we used two different methods of verifying that we did get the optimal value. Do not get too locked into one method of doing this verification that you forget about the other methods. Let’s work some another example that this time doesn’t involve a rectangle or box. Example 4 A manufacturer needs to make a cylindrical can that will hold 1.5 liters of liquid. Determine the dimensions of the can that will minimize the amount of material used in its construction. Solution In this problem the constraint is the volume and we want to minimize the amount of material used. This means that what we want to minimize is the surface area of the can and we’ll need to include both the walls of the can as well as the top and bottom “caps”. Here is a quick sketch to get us started off.


Calculus I


We’ll need the surface area of this can and that will be the surface area of the walls of the can (which is really just a cylinder) and the area of the top and bottom caps (which are just disks, and don’t forget that there are two of them). Note that if you think of a cylinder of height h and radius r as just a bunch of disks/circles of radius r stacked on top of each other the equations for the surface area and volume are pretty simple to remember. The volume is just the area of each of the disks times the height. Similarly, the surface area is just the circumference of the each circle times the height. The equations for the volume and surface area of a cylinder are then,

( )( ) ( )( )2 2 2 2V r h r h A r h rhπ π π π= = = =

Next, we’re also going to need the required volume in a better set of units than liters. Since we want length measurements for the radius and height we’ll need to use the fact that 1 Liter = 1000 cm3 to convert the 1.5 liters into 1500 cm3. This will in turn give a radius and height in terms of centimeters. Here are the equations that we’ll need for this problem and don’t forget that there two caps and so we’ll need the area from each.

2

2

Minimize : 2 2Constraint : 1500

A rh rr h

π π

π

= +

=

In this case it looks like our best option is to solve the constraint for h and plug this into the area function.

( ) 2 22 2

1500 1500 30002 2 2h A r r r rr r r

π π ππ π

= ⇒ = + = +

Notice that this formula will only make sense from a physical standpoint if 0r > which is a good thing as it is not defined at 0r = . Next, let’s get the first derivative.

( )3

2 2

3000 4 30004 rA r rr r

ππ

−′ = − =


Calculus I


From this we can see that we have two critical points : 0r = and 7503 6.2035r π= = . The first

critical point doesn’t make sense from a physical standpoint and so we can ignore that one. So we only have a single critical point to deal with here and notice that 6.2035 is the only value for which the derivative will be zero and hence the only place (with 0r > of course) that the derivative may change sign. It’s not difficult to check that if 6.2035r < then ( ) 0A r′ < and

likewise if 6.2035r > then ( ) 0A r′ > . The variant of the First Derivative Test above then tells

us that the absolute minimum value of the area (for 0r > ) must occur at 6.2035r = . All we need to do this is determine height of the can and we’ll be done.

( )2

1500 12.40706.2035

hπ

= =

Therefore if the manufacturer makes the can with a radius of 6.2035 cm and a height of 12.4070 cm the least amount of material will be used to make the can. As an interesting side problem and extension to the above example you might want to show that for a given volume, L, the minimum material will be used if 2h r= regardless of the volume of the can. In the examples to this point we’ve put in quite a bit of discussion in the solution. In the remaining problems we won’t be putting in quite as much discussion and leave it to you to fill in any missing details. Example 5 We have a piece of cardboard that is 14 inches by 10 inches and we’re going to cut out the corners as shown below and fold up the sides to form a box, also shown below. Determine the height of the box that will give a maximum volume.

Solution In this example, for the first time, we’ve run into a problem where the constraint doesn’t really have an equation. The constraint is simply the size of the piece of cardboard and has already been factored into the figure above. This will happen on occasion and so don’t get excited about it when it does. This just means that we have one less equation to worry about. In this case we


Calculus I


want to maximize the volume. Here is the volume, in terms of h and its first derivative. ( ) ( ) ( ) ( )2 3 214 2 10 2 140 48 4 140 96 12V h h h h h h h V h h h′= − − = − + = − +

Setting the first derivative equal to zero and solving gives the following two critical points,

12 39 1.9183, 6.0817

3h ±

= =

We now have an apparent problem. We have two critical points and we’ll need to determine which one is the value we need. In this case, this is easier than it looks. Go back to the figure in the problem statement and notice that we can quite easily find limits on h. The smallest h can be is 0h = even though this doesn’t make much sense as we won’t get a box in this case. Also from the 10 inch side we can see that the largest h can be is 5h = although again, this doesn’t make much sense physically. So, knowing that whatever h is it must be in the range 0 5h≤ ≤ we can see that the second critical point is outside this range and so the only critical point that we need to worry about is 1.9183. Finally, since the volume is defined and continuous on 0 5h≤ ≤ all we need to do is plug in the critical points and endpoints into the volume to determine which gives the largest volume. Here are those function evaluations.

( ) ( ) ( )0 0 1.9183 120.1644 5 0V V V= = =

So, if we take 1.9183h = we get a maximum volume. Example 6 A printer need to make a poster that will have a total area of 200 in2 and will have 1 inch margins on the sides, a 2 inch margin on the top and a 1.5 inch margin on the bottom. What dimensions will give the largest printed area? Solution This problem is a little different from the previous problems. Both the constraint and the function we are going to optimize are areas. The constraint is that the overall area of the poster must be 200 in2 while we want to optimize the printed area (i.e. the area of the poster with the margins taken out). Here is a sketch of the poster and we can see that once we’ve taken the margins into account the width of the printed area is 2w − and the height of the printer area is 3.5h − .


Calculus I


Here are the equations that we’ll be working with.

( ) ( )Maximize : 2 3.5Constraint : 200

A w hwh

= − −

=

Solving the constraint for h and plugging into the equation for the printed area gives,

( ) ( ) 200 4002 3.5 207 3.5A w w ww w

= − − = − −

The first and second derivatives are,

( ) ( )2

2 2 3

400 400 3.5 8003.5 wA w A ww w w

−′ ′′= − + = = −

From the first derivative we have the following three critical points. 4003.50 10.6904w w= = ± = ±

However, since we’re dealing with the dimensions of a piece of paper we know that we must have 0w > and so only 10.6904 will make sense. Also notice that provided 0w > the second derivative will always be negative and so in the range of possible optimal values of the width the area function is always concave down and so we know that the maximum printed area will be at 10.6904 inchesw = . The height of the paper that gives the maximum printed area is then,

200 18.7084 inches10.6904

h = =

We’ve worked quite a few examples to this point and we have quite a few more to work. However this section has gotten quite lengthy so let’s continue our examples in the next section. This is being done mostly because these notes are also being presented on the web and this will help to keep the load times on the pages down somewhat.


Calculus I


More Optimization Problems Because these notes are also being presented on the web we’ve broken the optimization examples up into several sections to keep the load times to a minimum. Do not forget the various methods for verifying that we have the optimal value that we looked at in the previous section. In this section we’ll just use them without acknowledging so make sure you understand them and can use them. So let’s get going on some more examples. Example 1 A window is being built and the bottom is a rectangle and the top is a semicircle. If there is 12 meters of framing materials what must the dimensions of the window be to let in the most light? Solution Okay, let’s ask this question again in slightly easier to understand terms. We want a window in the shape described above to have a maximum area (and hence let in the most light) and have a perimeter of 12 m (because we have 12 m of framing material). Little bit easier to understand in those terms. Here’s a sketch of the window. The height of the rectangular portion is h and because the semicircle is on top we can think of the width of the rectangular portion at 2r.

The perimeter (our constraint) is the lengths of the three sides on the rectangular portion plus half the circumference of a circle of radius r. The area (what we want to maximize) is the area of the rectangle plus half the area of a circle of radius r. Here are the equations we’ll be working with in this example.

21

2Maximize : 2Constraint : 12 2 2

A hr rh r r

ππ

= +

= + +

In this case we’ll solve the constraint for h and plug that into the area equation. ( ) ( ) 2 2 21 1 1 1

2 2 2 26 2 6 12 2h r r A r r r r r r r rπ π π π= − − ⇒ = − − + = − −

The first and second derivatives are,


Calculus I


( ) ( ) ( )12 4 4A r r A rπ π′ ′′= − + = − −

We can see that the only critical point is,

12 1.68034

rπ

= =+

We can also see that the second derivative is always negative (in fact it’s a constant) and so we can see that the maximum area must occur at this point. So, for the maximum area the semicircle on top must have a radius of 1.6803 and the rectangle must have the dimensions 3.3606 x 1.6803 (h x 2r). Example 2 Determine the area of the largest rectangle that can be inscribed in a circle of radius 4. Solution Huh? This problem is best described with a sketch. Here is what we’re looking for.

We want the area of the largest rectangle that we can fit inside a circle and have all of its corners touching the circle. To do this problem it’s easiest to assume that the circle (and hence the rectangle) is centered at the origin. Doing this we know that the equation of the circle will be 2 2 16x y+ = and that the right upper corner of the rectangle will have the coordinates ( ),x y . This means that

the width of the rectangle will be 2x and the height of the rectangle will be 2y. The area of the rectangle will then be, ( )( )2 2 4A x y xy= =


Calculus I


So, we’ve got the function we want to maximize (the area), but what is the constraint? Well since the coordinates of the upper right corner must be on the circle we know that x and y must satisfy the equation of the circle. In other words, the equation of the circle is the constraint. The first thing to do then is to solve the constraint for one of the variables.

216y x= ± − Since the point that we’re looking at is in the first quadrant we know that y must be positive and so we can take the “+” part of this. Plugging this into the area and computing the first derivative gives,

( )

( )

2

2 22

2 2

4 16

4 64 84 1616 16

A x x x

x xA x xx x

= −

−′ = − − =− −

Before getting the critical points let’s notice that we can limit x to the range 0 4x≤ ≤ since we are assuming that x is in the first quadrant and must stay inside the circle. Now the four critical points we get (two from the numerator and two from the denominator) are,

2

2

16 0 4

64 8 0 2 2

x x

x x

− = ⇒ = ±

− = ⇒ = ±

We only want critical points that are in the range of possible optimal values so that means that we

have two critical points to deal with : 2 2x = and 4x = . Notice however that the second critical point is also one of the endpoints of our interval. Now, area function is continuous and we have an interval of possible solution with finite endpoints so,

( ) ( ) ( )0 0 2 2 32 4 0A A A= = =

So, we can see that we’ll get the maximum area if 2 2x = and the corresponding value of y is,

( )216 2 2 8 2 2y = − = =

It looks like the maximum area will be found if the inscribed rectangle is in fact a square. We need to again make a point that was made several times in the previous section. We excluded several critical points in the work above. Do not always expect to do that. There will often be physical reasons to exclude zero and/or negative critical points, however, there will be problems where these are perfectly acceptable values. You should always write down every possible critical point and then exclude any that can’t be possible solutions. This keeps you in the habit of


Calculus I


finding all the critical points and then deciding which ones you actually need and that in turn will make it less likely that you’ll miss one when it is actually needed. Example 3 Determine the point(s) on 2 1y x= + that are closest to (0,2). Solution Here’s a quick sketch of the situation.

So, we’re looking for the shortest length of the dashed line. Notice as well that if the shortest distance isn’t at 0x = there will be two points on the graph, as we’ve shown above, that will give the shortest distance. This is because the parabola is symmetric to the y-axis and the point in question is on the y-axis. This won’t always be the case of course so don’t always expect two points in these kinds of problems. In this case we need to minimize the distance between the point (0,2) and any point that is one the graph (x,y). Or,

( ) ( ) ( )2 2 220 2 2d x y x y= − + − = + − If you think about the situation here it makes sense that the point that minimizes the distance will also minimize the square of the distance and so since it will be easier to work with we will use the square of the distance and minimize that. If you aren’t convinced of this we’ll take a closer look at this after this problem. So, the function that we’re going to minimize is,

( )22 2 2D d x y= = + − The constraint in this case is the function itself since the point must lie on the graph of the function. At this point there are two methods for proceeding. One of which will require significantly more work than the other. Let’s take a look at both of them.


Calculus I


Solution 1 In this case we will use the constraint in probably the most obvious way. We already have the constraint solved for y so let’s plug that into the square of the distance and get the derivatives.

( ) ( )( ) ( )( )

22 2 4 2

3 2

2

1 2 1

4 2 2 2 1

12 2

D x x x x x

D x x x x x

D x x

= + + − = − +

′ = − = −

′′ = −

So, it looks like there are three critical points for the square of the distance and notice that this time, unlike pretty much every previous example we’ve worked, we can’t exclude zero or negative numbers. They are perfectly valid possible optimal values this time.

10,2

x x= = ±

Before going any farther, let’s check these in the second derivative to see if they are all relative minimums.

( ) 1 10 2 0 4 42 2

D D D ′′ ′′ ′′= − < = − =

So, 0x = is a relative maximum and so can’t possibly be the minimum distance. That means that we’ve got two critical points. The question is how do we verify that these give the minimum distance and yes we did mean to say that both will give the minimum distance. Recall from our sketch above that if x gives the minimum distance then so will –x and so if gives the minimum distance then the other should as well. None of the methods we discussed in the previous section will really work here. We don’t have an interval of possible solutions with finite endpoints and both the first and second derivative change sign. In this case however, we can still verify that they are the points that give the minimum distance.

First, notice that if we are working on the interval 1 12 2

, − then the endpoints of this interval

(which are also the critical points) are in fact where the absolute minimum of the function occurs in this interval. Next we can see that if 1

2x < − then ( ) 0D x′ < . Or in other words, if 1

2x < − the function is

decreasing until it hits 12

x = − and so must always be larger than the function at 12

x = − .

Similarly, 1

2x > then ( ) 0D x′ > and so the function is always increasing to the right of


Calculus I


12

x = − and so must be larger than the function at 12

x = − .

So, putting all of this together tells us that we do in fact have an absolute minimum at 1

2x = ± .

All that we need to do is to find the value of y for these points.

1 3:22

1 3:22

x y

x y

= =

= − =

So, the points on the graph that are closest to (0,2) are,

1 3 1 3, ,2 22 2

−

Solution 2 The first solution that we worked was actually the long solution. There is a much shorter solution to this problem. Instead of plugging y into the square of the distance let’s plug in x. From the constraint we get, 2 1x y= − and notice that the only place x show up in the square of the distance it shows up as x2 and let’s just plug this into the square of the distance. Doing this gives,

( ) ( )( )( )

2 21 2 3 3

2 3

2

D y y y y y

D y y

D y

= − + − = − +

′ = −

′′ =

There is now a single critical point, 3

2y = , and since the second derivative is always positive we know that this point must give the absolute minimum. So all that we need to do at this point is find the value(s) of x that go with this value of y.

2 3 1 112 2 2

x x= − = ⇒ = ±

The points are then,

1 3 1 3, ,2 22 2

−

So, for significantly less work we got exactly the same answer. This previous example had a couple of nice points. First, as pointed out in the problem, we couldn’t exclude zero or negative critical points this time as we’ve done in all the previous


Calculus I


examples. Again, be careful to not get into the habit of always excluding them as we do many of the examples we’ll work. Next, some of these problems will have multiple solution methods and sometimes one will be significantly easier than the other. The method you use is up to you and often the difficulty of any particular method is dependent upon the person doing the problem. One person may find one way easier and other person may find a different method easier. Finally, as we saw in the first solution method sometimes we’ll need to use a combination of the optimal value verification methods we discussed in the previous section. Now, before we move onto the next example let’s take a look at the claim above that we could find the location of the point that minimizes the distance by finding the point that minimizes the square of the distance. We’ll generalize things a little bit, Fact Suppose that we have a positive function, ( ) 0f x > , that exists everywhere then ( )f x and

( ) ( )g x f x= will have the same critical points and the relative extrema will occur at the same

points. This is simple enough to prove so let’s do that here. First let’s take the derivative of ( )g x and

see what we can determine about the critical points of ( )g x .

( ) ( ) ( ) ( )( )

11 22 2

f xg x f x f x

f x− ′

′ ′= =

Let’s plug x c= into this to get,

( ) ( )( )2

f cg c

f c

′′ =

By assumption we know that ( )f c exists and ( ) 0f c > and therefore the denominator of this

will always exist and will never be zero. We’ll need this in several places so we can’t forget this. If ( ) 0f c′ = then because we know that the denominator will not be zero here we must also have

( ) 0g c′ = . Likewise, if ( ) 0g c′ = then we must have ( ) 0f c′ = . So, ( )f x and ( )g x will

have the same critical points in which the derivatives will be zero. Next, if ( )f c′ doesn’t exist then ( )g c′ will also not exist and likewise if ( )g c′ doesn’t exist

then because we know that the denominator will not be zero then this means that ( )f c′ will also


Calculus I


not exist. Therefore, ( )f x and ( )g x will have the same critical points in which the derivatives

does not exist. So, the upshot of all this is that ( )f x and ( )g x will have the same critical points.

Next, let’s notice that because we know that ( )2 0f x > then ( )f x′ and ( )g x′ will have the

same sign and so if we apply the first derivative test (and recalling that they have the same critical points) to each of these functions we can see that the results will be the same and so the relative extrema will occur at the same points. Note that we could also use the second derivative test to verify that the critical points will have the same classification if we wanted to. The second derivative is (and you should see if you can use the quotient rule to verify this),

( ) ( ) ( ) ( ) ( )( )

1222

4f x f x f x f x

g xf x

−′′ ′− ′′ =

Then if x c= is a critical point such that ( ) 0f c′ = (and so we can use the second derivative

test) we get,

( ) ( ) ( ) ( ) ( )

( )

( ) ( ) [ ] ( )( )

( ) ( )( )

122

12 2

24

2 04 2

f c f c f c f cg c

f c

f c f c f c f c f cf c f c

−

−

′′ ′− ′′ =

′′ ′′− = =

Now, because we know that ( )2 0f c > and by assumption ( ) 0f c > we can see that ( )f c′′

and ( )g c′′ will have the same sign and so will have the same conclusion from the second

derivative test. So, now that we have that out of the way let’s work some more examples. Example 4 A 2 feet piece of wire is cut into two pieces and once piece is bent into a square and the other is bent into an equilateral triangle. Where should the wire cut so that the total area enclosed by both is minimum and maximum? Solution Before starting the solution recall that an equilateral triangle is a triangle with three equal sides and each of the interior angles are 3

π (or 60° ).


Calculus I


Now, this is another problem where the constraint isn’t really going to be given by an equation, it is simply that there is 2 ft of wire to work with and this will be taken into account in our work. So, let’s cut the wire into two pieces. The first piece will have length x which we’ll bend into a square and each side will have length 4

x . The second piece will then have length 2 x− (we just used the constraint here…) and we’ll bend this into an equilateral triangle and each side will have length ( )1

3 2 x− . Here is a sketch of all this.

As noted in the sketch above we also will need the height of the triangle. This is easy to get if you realize that the dashed line divides the equilateral triangle into two other triangles. Let’s look at the right one. The hypotenuse is ( )1

3 2 x− while the lower right angle is 3π . Finally the

height is then the opposite side to the lower right angle so using basic right triangle trig we arrive at the height of the triangle as follows.

( ) ( ) ( ) ( ) ( ) ( )3 31 13 3 3 3 2 6sin 2 sin 2 2opp

hyp opp x x xπ π= ⇒ = − = − = −

So, the total area of both objects is then,

( ) ( ) ( )( ) ( )( ) ( )22 23 31 14 2 3 6 16 362 2 2xxA x x x x= + − − = + −

Here’s the first derivative of the area.

( ) ( )( )( )3 3 38 36 8 9 182 2 1x xA x x x′ = + − − = − +

Setting this equal to zero and solving gives the single critical point of,

8 39 4 3

0.8699x+

= =

Now, let’s notice that the problem statement asked for both the minimum and maximum enclosed area and we got a single critical point. This clearly can’t be the answer to both, but this is not the problem that it might seem to be.


Calculus I


Let’s notice that x must be in the range 0 2x≤ ≤ and since the area function is continuous we use the basic process for finding absolute extrema of a function.

( ) ( ) ( )0 0.1925 0.8699 0.1087 2 0.25A A A= = =

So, it looks like the minimum area will arise if we take 0.8699x = while the maximum area will arise if we take the whole piece of wire and bend it into a square. As the previous problem illustrated we can’t get too locked into the answers always occurring at the critical points as they have to this point. That will often happen, but one of the extrema in the previous problem was at an endpoint and that will happen on occasion. Example 5 A piece of pipe is being carried down a hallway that is 10 feet wide. At the end of the hallway the there is a right-angled turn and the hallway narrows down to 8 feet wide. What is the longest pipe that can be carried (always keeping it horizontal) around the turn in the hallway? Solution Let’s start off with a sketch of the situation so we can get a grip on what’s going on and how we’re going to have to go about solving this.

The largest pipe that can go around the turn will do so in the position shown above. One end will be touching the outer wall of the hall way at A and C and the pipe will touch the inner corner at B. Let’s assume that the length of the pipe in the small hallway is 1L while 2L is the length of the

pipe in the large hallway. The pipe then has a length of 1 2L L L= + .

Now, if 0θ = then the pipe is completely in the wider hallway and we can see that as 0θ → then L → ∞ . Likewise, if 2

πθ = the pipe is completely in the narrow hallway and as 2πθ →

we also have L → ∞ . So, somewhere in the interval 20 πθ< < is an angle that will minimize L and oddly enough that is the length that we’re after. The largest pipe that will fit around the turn will in fact be the minimum value of L. The constraint for this problem is not so obvious and there are actually two of them. The constraints for this problem are the widths of the hallways. We’ll use these to get an equation for


Calculus I


L in terms of θ and then we’ll minimize this new equation. So, using basic right triangle trig we can see that, 1 28sec 10csc 8sec 10cscL L Lθ θ θ θ= = ⇒ = +

So, differentiating L gives, 8sec tan 10csc cotL θ θ θ θ′ = − Setting this equal to zero and solving gives,

23

8sec tan 10csc cotsec tan 10csc cot 8

sin tan 5 tan 1.25cos 4

θ θ θ θθ θθ θ

θ θθ

θ

=

=

= ⇒ =

Solving for θ gives,

( )13 3tan 1.25 tan 1.25 0.8226θ θ −= ⇒ = =

So, if 0.8226θ = radians then the pipe will have a minimum length and will just fit around the turn. Anything larger will not fit around the turn and so the largest pipe that can be carried around the turn is, ( ) ( )8sec 0.8226 10csc 0.8226 25.4033 feetL = + =

Example 6 Two poles, one 6 meters tall and one 15 meters tall, are 20 meters apart. A length of wire is attached to the top of each pole and it is also staked to the ground somewhere between the two poles. Where should the wire be staked so that the minimum amount of wire is used? Solution As always let’s start off with a sketch of this situation.


Calculus I


The total length of the wire is 1 2L L L= + and we need to determine the value of x that will

minimize this. The constraint in this problem is that the poles must be 20 meters apart and that x must be in the range 0 20x≤ ≤ . The first thing that we’ll need to do here is to get the length of wire in terms of x, which is fairly simple to do using the Pythagorean Theorem.

( )22 2 21 236 225 20 36 625 40L x L x L x x x= + = + − = + + − +

Not the nicest function we’ve had to work with but there it is. Note however, that it is a continuous function and we’ve got an interval with finite endpoints and so finding the absolute minimum won’t require much more work than just getting the critical points of this function. So, let’s do that. Here’s the derivative.

2 2

2036 625 40

x xLx x x

−′ = ++ − +

Setting this equal to zero gives,

( )

2 2

2 2

20 036 625 40

625 40 20 36

x xx x x

x x x x x

−+ =

+ − +

− + = − − +

It’s probably been quite a while since you’ve been asked to solve something like this. To solve this we’ll need to square both sides to get rid of the roots, but this will cause problems as well soon see. Let’s first just square both sides and solve that equation.

( ) ( ) ( )

( )( )

22 2 2

2 3 4 2 3 4

2

40 403 7

625 40 20 36

625 40 14400 1440 436 40189 1440 14400 0

9 3 40 7 40 0 ,

x x x x x

x x x x x x xx x

x x x x

− + = − +

− + = − + − +

+ − =

+ − = ⇒ = − =

Note that if you can’t do that factoring done worry, you can always just use the quadratic formula and you’ll get the same answers. Okay two issues that we need to discuss briefly here. The first solution above (note that I didn’t call it a critical point…) doesn’t make any sense because it is negative and outside of the range of possible solutions and so we can ignore it. Secondly, and maybe more importantly, if you were to plug 40

3x = − into the derivative you would not get zero and so is not even a critical point. How is this possible? It is a solution after all. We’ll recall that we squared both sides of the equation above and it was mentioned at the


Calculus I


time that this would cause problems. We’ll we’ve hit those problems. In squaring both sides we’ve inadvertently introduced a new solution to the equation. When you do something like this you should ALWAYS go back and verify that the solutions that you get are in fact solutions to the original equation. In this case we were lucky and the “bad” solution also happened to be outside the interval of solutions we were interested in but that won’t always be the case. So, if we go back and do a quick verification we can in fact see that the only critical point is

407 5.7143x = = and this is nicely in our range of acceptable solutions.

Now all that we need to do is plug this critical point and the endpoints of the wire into the length formula and identify the one that gives the minimum value. ( ) ( ) ( )40

70 31 29 20 35.8806L L L= = =

So, we will get the minimum length of wire if we stake it to the ground 40

7 feet from the smaller pole. Let’s do a modification of the above problem that asks a completely different question. Example 7 Two poles, one 6 meters tall and one 15 meters tall, are 20 meters apart. A length of wire is attached to the top of each pole and it is also staked to the ground somewhere between the two poles. Where should the wire be staked so that the angle formed by the two pieces of wire at the stake is a maximum? Solution Here’s a sketch for this example.

The equation that we’re going to need to work with here is not obvious. Let’s start with the following fact.

180δ θ ϕ π+ + = = Note that we need to make sure that the equation is equal to π because of how we’re going to work this problem. Now, basic right triangle trig tells us the following,


Calculus I


( )( )

16 6

115 1520 20

tan tan

tan tanx x

x x

δ δ

ϕ ϕ

−

−− −

= ⇒ =

= ⇒ =

Plugging these into the equation above and solving for θ gives, ( ) ( )1 16 15

20tan tan xxθ π − −−= − −

Note that this is the reason for the π in our equation. The inverse tangents give angles that are in radians and so can’t use the 180 that we’re used to in this kind of equation. Next we’ll need the derivative so hopefully you’ll recall how to differentiate inverse tangents.

( ) ( ) ( )

( )( )

( )( )

2 2 226 1520

22

2

2 2 2 2

1 6 1 15201 1

6 1536 20 225

3 3 80 10706 1536 40 625 36 40 625

xx x x

x x

x xx x x x x x

θ−

′ = − − − −+ +

= −+ − +

− + −= − =

+ − + + − +

Setting this equal to zero and solving give the following two critical points.

40 48103 36.4514, 9.7847x − ±= = −

The first critical point is not in the interval of possible solutions and so we can exclude it. It’s not difficult to show that if 0 9.7847x≤ ≤ that 0θ ′ > and if 9.7847 20x≤ ≤ that 0θ ′ < and so when 9.7847x = we will get the maximum value of θ . Example 8 A trough for holding water is be formed by taking a piece of sheet metal 60 cm wide and folding the 20 cm on either end up as shown below. Determine the angle θ that will maximize the amount of water that the trough can hold.

Solution Now, in this case we are being asked to maximize the volume that a trough can hold, but if you think about it the volume of a trough in this shape is nothing more than the cross-sectional area times the length of the trough. So for a given length in order to maximize the volume all you really need to do is maximize the cross-sectional area. To get a formula for the cross-sectional area let’s redo the sketch above a little.


Calculus I


We can think of the cross-sectional area as a rectangle in the middle with width 20 and height h and two identical triangles on either end with height h, base b and hypotenuse 20. Also note that basic geometry tells us that the angle between the hypotenuse and the base must also be the same angle θ that we had in our original sketch. Also, basic right triangle trig tells us that the base and height can be written as, 20cos 20sinb hθ θ= = The cross-sectional area for the whole trough, in terms of θ , is then,

( ) ( )( ) ( )1220 2 400sin 20cos 20sin 400 sin sin cosA h bh θ θ θ θ θ θ= + = + = +

The derivative of the area is,

( ) ( )( )( )

( )( )( )

2 2

2 2

2

400 cos cos sin

400 cos cos 1 cos

400 2cos cos 1

400 2cos 1 cos 1

A θ θ θ θ

θ θ θ

θ θ

θ θ

′ = + −

= + − −

= + −

= − +

So, we have either,

12 32cos 1 0 cos

cos 1 0 cos 1

πθ θ θθ θ θ π

− = ⇒ = ⇒ =

+ = ⇒ = − ⇒ =

However, we can see that θ must be in the interval 20 πθ≤ ≤ or we won’t get a trough in the proper shape. Therefore, the second critical point makes no sense and also note that we don’t need to add on the standard “ 2 nπ+ ” for the same reason. Finally, since the equation for the area is continuous all we need to do is plug in the critical point and the end points to find the one that gives the maximum area. ( ) ( ) ( )3 20 0 519.6152 400A A Aπ π= = =

So, we will get a maximum cross-sectional area, and hence a maximum volume, when 3

πθ = .


Calculus I


Indeterminate Forms and L’Hospital’s Rule Back in the chapter on Limits we saw methods for dealing with the following limits.

2 2

24

16 4 5lim lim4 1 3x x

x x xx x→ →∞

− −− −

In the first limit if we plugged in 4x = we would get 0/0 and in the second limit if we “plugged” in infinity we would get ∞ −∞ (recall that as x goes to infinity a polynomial will behave in the same fashion that its largest power behaves). Both of these are called indeterminate forms. In both of these cases there are competing interests or rules and it’s not clear which will win out. In the case of 0/0 we typically think of a fraction that has a numerator of zero as being zero. However, we also tend to think of fractions in which the denominator is going to zero as infinity or might not exist at all. Likewise, we tend to think of a fraction in which the numerator and denominator are the same as one. So, which will win out? Or will neither win out and they all “cancel out” and the limit will reach some other value? In the case of ∞ −∞ we have a similar set of problems. If the numerator of a fraction is going to infinity we tend to think of the whole fraction going to infinity. Also if the denominator is going to infinity we tend to think of the fraction as going to zero. We also have the case of a fraction in which the numerator and denominator are the same (ignoring the minus sign) and so we might get -1. Again, it’s not clear which of these will win out, if any of them will win out. With the second limit there is the further problem that infinity isn’t really a number and so we really shouldn’t even treat it like a number. Much of the time it simply won’t behave as we would expect it to if it was a number. To look a little more into this check out the Types of Infinity section in the Extras chapter at the end of this document. This is the problem with indeterminate forms. It’s just not clear what is happening in the limit. There are other types of indeterminate forms as well. Some other types are, ( )( ) 0 00 1 0∞± ∞ ∞ ∞ − ∞ These all have competing interests or rules that tell us what should happen and it’s just not clear which, if any, of the interests or rules will win out. The topic of this section is how to deal with these kinds of limits. As already pointed out we do know how to deal with some kinds of indeterminate forms already. For the two limits above we work them as follows.

( )2

4 4

16lim lim 4 84x x

x xx→ →

−= + =

−


Calculus I


2

2

2

544 5 4lim lim 11 3 33x x

x x xx

x→∞ →∞

−−= = −

− −

In the first case we simply factored, canceled and took the limit and in the second case we factored out an 2x from both the numerator and the denominator and took the limit. Notice as well that none of the competing interests or rules in these cases won out! That is often the case. So we can deal with some of these. However what about the following two limits.

20

sinlim limx

x x

xx x→ →∞

e

This first is a 0/0 indeterminate form, but we can’t factor this one. The second is an ∞ ∞

indeterminate form, but we can’t just factor an 2x out of the numerator. So, nothing that we’ve got in our bag of tricks will work with these two limits. This is where the subject of this section comes into play. L’Hospital’s Rule Suppose that we have one of the following cases,

( )( )

( )( )

0lim OR lim0x a x a

f x f xg x g x→ →

± ∞= =

± ∞

where a can be any real number, infinity or negative infinity. In these cases we have,

( )( )

( )( )

lim limx a x a

f x f xg x g x→ →

′=

′

So, L’Hospital’s Rule tells us that if we have an indeterminate form 0/0 or ∞ ∞ all we need to do is differentiate the numerator and differentiate the denominator and then take the limit. Before proceeding with examples let me address the spelling of “L’Hospital”. The more modern spelling is “L’Hôpital”. However, when I first learned Calculus my teacher used the spelling that I use in these notes and the first text book that I taught Calculus out of also used the spelling that I use here. Also, as noted on the Wikipedia page for L’Hospital,

“In the 17th and 18th centuries, the name was commonly spelled "l'Hospital", however, French spellings have been altered: the silent 's' has been dropped and replaced with the circumflex over the preceding vowel. The former spelling is still used in English where there is no circumflex.”


Calculus I


So, the spelling that I’ve used here is an acceptable spelling of his name, albeit not the modern spelling, and because I’m used to spelling it as “L’Hospital” that is the spelling that I’m going to use in these notes. Let’s work some examples. Example 1 Evaluate each of the following limits.

(a) 0

sinlimx

xx→

[Solution]

(b) 4 2

31

5 4 1lim10 9t

t tt t→

− −− −

[Solution]

(c) 2limx

x x→∞

e [Solution]

Solution

(a) 0

sinlimx

xx→

So, we have already established that this is a 0/0 indeterminate form so let’s just apply L’Hospital’s Rule.

0 0

sin cos 1lim lim 11 1x x

x xx→ →

= = =


(b) 4 2

31

5 4 1lim10 9t

t tt t→

− −− −

In this case we also have a 0/0 indeterminate form and if we were really good at factoring we could factor the numerator and denominator, simplify and take the limit. However, that’s going to be more work than just using L’Hospital’s Rule.

4 2 3

3 21 1

5 4 1 20 8 20 8 3lim lim10 9 1 27 1 27 7t t

t t t tt t t→ →

− − − −= = = −

− − − − − −


(c) 2limx

x x→∞

e

This was the other limit that we started off looking at and we know that it’s the indeterminate form ∞ ∞ so let’s apply L’Hospital’s Rule.

2lim lim2

x x

x xx x→∞ →∞=

e e

Now we have a small problem. This new limit is also a ∞ ∞ indeterminate form. However, it’s


Calculus I


not really a problem. We know how to deal with these kinds of limits. Just apply L’Hospital’s Rule.

2lim lim lim2 2

x x x

x x xx x→∞ →∞ →∞= = = ∞

e e e

Sometimes we will need to apply L’Hospital’s Rule more than once.

[Return to Problems] L’Hospital’s Rule works great on the two indeterminate forms 0/0 and ± ∞ ± ∞ . However, there are many more indeterminate forms out there as we saw earlier. Let’s take a look at some of those and see how we deal with those kinds of indeterminate forms. We’ll start with the indeterminate form ( )( )0 ± ∞ .

Example 2 Evaluate the following limit.

0lim lnx

x x+→

Solution Note that we really do need to do the right-hand limit here. We know that the natural logarithm is only defined for positive x and so this is the only limit that makes any sense. Now, in the limit, we get the indeterminate form ( ) ( )0 −∞ . L’Hospital’s Rule won’t work on

products, it only works on quotients. However, we can turn this into a fraction if we rewrite things a little.

0 0

lnlim ln lim 1x x

xx xx

+ +→ →=

The function is the same, just rewritten, and the limit is now in the form −∞ ∞ and we can now use L’Hospital’s Rule.

0 0 0

2

1lnlim ln lim lim1 1x x x

x xx xx x

+ + +→ → →= =

−

Now, this is a mess, but it cleans up nicely.

( )0 0 0

2

1lim ln lim lim 01x x x

xx x xx

+ + +→ → →= = − =

−

In the previous example we used the fact that we can always write a product of functions as a quotient by doing one of the following.


Calculus I


( ) ( ) ( )

( )( ) ( ) ( )

( )OR1 1

g x f xf x g x f x g x

f x g x= =

Using these two facts will allow us to turn any limit in the form ( )( )0 ± ∞ into a limit in the form

0/0 or ± ∞ ± ∞ . Which one of these two we get after doing the rewrite will depend upon which

fact we used to do the rewrite. One of the rewrites will give 0/0 and the other will give ± ∞ ± ∞ . It all depends on which function stays in the numerator and which gets moved down to the denominator. Let’s take a look at another example. Example 3 Evaluate the following limit. lim x

xx

→−∞e

Solution So, it’s in the form ( )( )0∞ . This means that we’ll need to write it as a quotient. Moving the x

to the denominator worked in the previous example so let’s try that with this problem as well.

lim lim 1

xx

x xx

x→−∞ →−∞

=ee

Writing the product in this way gives us a product that has the form 0/0 in the limit. So, let’s use L’Hospital’s Rule on the quotient.

2 3 4

lim lim lim lim lim1 1 2 6

x x x xx

x x x x xx

x x x x→−∞ →−∞ →−∞ →−∞ →−∞

= = = = =− −

e e e ee L

Hummmm…. This doesn’t seem to be getting us anywhere. With each application of L’Hospital’s Rule we just end up with another 0/0 indeterminate form and in fact the derivatives seem to be getting worse and worse. Also note that if we simplified the quotient back into a product we would just end up with either ( )( )0∞ or ( )( )0−∞ and so that won’t do us any

good. This does not mean however that the limit can’t be done. It just means that we moved the wrong function to the denominator. Let’s move the exponential function instead.

lim lim lim1x

xx x xx

x xx −→−∞ →−∞ →−∞= =e

ee

Note that we used the fact that,

1 xx

−= ee

to simplify the quotient up a little. This will help us when it comes time to take some derivatives.


Calculus I


The quotient is now an indeterminate form of −∞ ∞ and use L’Hospital’s Rule gives,

1lim lim lim 0xx xx x x

xx − −→−∞ →−∞ →−∞= = =

−e

e e

So, when faced with a product ( )( )0 ± ∞ we can turn it into a quotient that will allow us to use

L’Hospital’s Rule. However, as we saw in the last example we need to be careful with how we do that on occasion. Sometimes we can use either quotient and in other cases only one will work. Let’s now take a look at the indeterminate forms, 0 01 0∞ ∞ These can all be dealt with in the following way so we’ll just work one example. Example 4 Evaluate the following limit.

1

lim xx

x→∞

Solution In the limit this is the indeterminate form 0∞ . We’re actually going to spend most of this problem on a different limit. Let’s first define the following.

1xy x=

Now, if we take the natural log of both sides we get,

( )1 1 lnln ln lnx xy x x

x x

= = =

Let’s now take a look at the following limit.

( )1lnlim ln lim lim 01x x x

x xyx→∞ →∞ →∞

= = =

This limit was just a L’Hospital’s Rule problem and we know how to do those. So, what did this have to do with our limit? Well first notice that,

( )ln y y=e and so our limit could be written as,

( )1

lnlim lim lim yxx x x

x y→∞ →∞ →∞

= = e

We can now use the limit above to finish this problem.

( ) ( )1lim lnln 0lim lim lim 1x

yyxx x x

x y →∞

→∞ →∞ →∞= = = = =e e e


Calculus I


With L’Hospital’s Rule we are now able to take the limit of a wide variety of indeterminate forms that we were unable to deal with prior to this section.


Calculus I


Linear Approximations In this section we’re going to take a look at an application not of derivatives but of the tangent line to a function. Of course, to get the tangent line we do need to take derivatives, so in some way this is an application of derivatives as well. Given a function, ( )f x , we can find its tangent at x a= . The equation of the tangent line,

which we’ll call ( )L x for this discussion, is,

( ) ( ) ( )( )L x f a f a x a′= + − Take a look at the following graph of a function and its tangent line.

From this graph we can see that near x a= the tangent line and the function have nearly the same graph. On occasion we will use the tangent line, ( )L x , as an approximation to the

function, ( )f x , near x a= . In these cases we call the tangent line the linear approximation to

the function at x a= . So, why would we do this? Let’s take a look at an example.

Example 1 Determine the linear approximation for ( ) 3f x x= at 8x = . Use the linear

approximation to approximate the value of 3 8.05 and 3 25 . Solution Since this is just the tangent line there really isn’t a whole lot to finding the linear approximation.

( ) ( ) ( )23

3 2

1 1 18 2 83 123

f x x f fx

−′ ′= = = =


Calculus I


The linear approximation is then,

( ) ( )1 1 42 812 12 3

L x x x= + − = +

Now, the approximations are nothing more than plugging the given values of x into the linear approximation. For comparison purposes we’ll also compute the exact values.

( )

( )

3

3

8.05 2.00416667 8.05 2.00415802

25 3.41666667 25 2.92401774

L

L

= =

= =

So, at 8.05x = this linear approximation does a very good job of approximating the actual value. However, at 25x = it doesn’t do such a good job. This shouldn’t be too surprising if you think about. Near 8x = both the function and the linear approximation have nearly the same slope and since they both pass through the point ( )8, 2 they

should have nearly the same value as long as we stay close to 8x = . However, as we move away from 8x = the linear approximation is a line and so will always have the same slope while the function’s slope will change as x changes and so the function will, in all likelihood, move away from the linear approximation. Here’s a quick sketch of the function and its linear approximation at 8x = .

As noted above, the farther from 8x = we get the more distance separates the function itself and its linear approximation. Linear approximations do a very good job of approximating values of ( )f x as long as we stay

“near” x a= . However, the farther away from x a= we get the worse the approximation is liable to be. The main problem here is that how near we need to stay to x a= in order to get a good approximation will depend upon both the function we’re using and the value of x a= that


Calculus I


we’re using. Also, there will often be no easy way of predicting how far away from x a= we can get and still have a “good” approximation. Let’s take a look at another example that is actually used fairly heavily in some places. Example 2 Determine the linear approximation for sinθ at 0θ = . Solution Again, there really isn’t a whole lot to this example. All that we need to do is compute the tangent line to sinθ at 0θ = .

( ) ( )( ) ( )

sin cos

0 0 0 1

f f

f f

θ θ θ θ′= =

′= =

The linear approximation is,

( ) ( ) ( )( )

( )( )0 0

0 1 0

L f f aθ θ

θ

θ

′= + −

= + −

=

So, as long as θ stays small we can say that sinθ θ≈ . This is actually a somewhat important linear approximation. In optics this linear approximation is often used to simplify formulas. This linear approximation is also used to help describe the motion of a pendulum and vibrations in a string.


Calculus I


Differentials In this section we’re going to introduce a notation that we’ll be seeing quite a bit in the next chapter. We will also look at an application of this new notation. Given a function ( )y f x= we call dy and dx differentials and the relationship between them is

given by, ( )dy f x dx′= Note that if we are just given ( )f x then the differentials are df and dx and we compute them in

the same manner. ( )df f x dx′= Let’s compute a couple of differentials. Example 1 Compute the differential for each of the following.

(a) 3 24 7y t t t= − + (b) ( )2 sin 2w x x=

(c) ( ) 43 zf z −= e Solution Before working any of these we should first discuss just what we’re being asked to find here. We defined two differentials earlier and here we’re being asked to compute a differential. So, which differential are we being asked to compute? In this kind of problem we’re being asked to compute the differential of the function. In other words, dy for the first problem, dw for the second problem and df for the third problem. Here are the solutions. Not much to do here other than take a derivative and don’t forget to add on the second differential to the derivative.

(a) ( )23 8 7dy t t dt= − +

(b) ( ) ( )( )22 sin 2 2 cos 2dw x x x x dx= +

(c) 43 34 zdf z dz−= − e


Calculus I


There is a nice application to differentials. If we think of x∆ as the change in x then

( ) ( )y f x x f x∆ = + ∆ − is the change in y corresponding to the change in x. Now, if x∆ is

small we can assume that y dy∆ ≈ . Let’s see an illustration of this idea.

Example 2 Compute dy and y∆ if ( )2cos 1y x x= + − as x changes from 2x = to 2.03x = . Solution First let’s compute actual the change in y, y∆ .

( )( ) ( )( )2 2cos 2.03 1 2.03 cos 2 1 2 0.083581127y∆ = + − − + − =

Now let’s get the formula for dy.

( )( )22 sin 1 1dy x x dx= − + − Next, the change in x from 2x = to 2.03x = is 0.03x∆ = and so we then assume that

0.03dx x≈ ∆ = . This gives an approximate change in y of,

( ) ( )( )( )22 2 sin 2 1 1 0.03 0.085070913dy = − + − = We can see that in fact we do have that y dy∆ ≈ provided we keep x∆ small. We can use the fact that y dy∆ ≈ in the following way. Example 3 A sphere was measured and its radius was found to be 45 inches with a possible error of no more that 0.01 inches. What is the maximum possible error in the volume if we use this value of the radius? Solution First, recall the equation for the volume of a sphere.

343

V rπ=

Now, if we start with 45r = and use 0.01dr r≈ ∆ = then V dV∆ ≈ should give us maximum error. So, first get the formula for the differential. 24dV r drπ= Now compute dV.

( ) ( )2 34 45 0.01 254.47 inV dV π∆ ≈ = =


Calculus I


The maximum error in the volume is then approximately 254.47 in3. Be careful to not assume this is a large error. On the surface it looks large, however if we compute the actual volume for 45r = we get 3381,703.51 inV = . So, in comparison the error in the volume is,

254.47 100 0.067%

381703.51× =

That’s not much possible error at all!


Calculus I


Newton’s Method The next application that we’ll take a look at in this chapter is an important application that is used in many areas. If you’ve been following along in the chapter to this point it’s quite possible that you’ve gotten the impression that many of the applications that we’ve looked at are just made up by us to make you work. This is unfortunate because all of the applications that we’ve looked at to this point are real applications that really are used in real situations. The problem is often that in order to work more meaningful examples of the applications we would need more knowledge than we generally have about the science and/or physics behind the problem. Without that knowledge we’re stuck doing some fairly simplistic examples that often don’t seem very realistic at all and that makes it hard to understand that the application we’re looking at is a real application. That is going to change in this section. This is an application that we can all understand and we can all understand needs to be done on occasion even if we don’t understand the physics/science behind an actual application. In this section we are going to look at a method for approximating solutions to equations. We all know that equations need to be solved on occasion and in fact we’ve solved quite a few equations ourselves to this point. In all the examples we’ve looked at to this point we were able to actually find the solutions, but it’s not always possible to do that exactly and/or do the work by hand. That is where this application comes into play. So, let’s see what this application is all about. Let’s suppose that we want to approximate the solution to ( ) 0f x = and let’s also suppose that

we have somehow found an initial approximation to this solution say, x0. This initial approximation is probably not all that good and so we’d like to find a better approximation. This is easy enough to do. First we will get the tangent line to ( )f x at x0.

( ) ( ) ( )0 0 0y f x f x x x′= + − Now, take a look at the graph below.


Calculus I


The blue line (if you’re reading this in color anyway…) is the tangent line at x0. We can see that this line will cross the x-axis much closer to the actual solution to the equation than x0 does. Let’s call this point where the tangent at x0 crosses the x-axis x1 and we’ll use this point as our new approximation to the solution. So, how do we find this point? Well we know it’s coordinates, ( )1,0x , and we know that it’s on

the tangent line so plug this point into the tangent line and solve for x1 as follows,

( ) ( ) ( )( )( )

( )( )

0 0 1 0

01 0

0

01 0

0

0 f x f x x x

f xx x

f x

f xx x

f x

′= + −

− = −′

= −′

So, we can find the new approximation provided the derivative isn’t zero at the original approximation. Now we repeat the whole process to find an even better approximation. We form up the tangent line to ( )f x at x1 and use its root, which we’ll call x2, as a new approximation to the actual

solution. If we do this we will arrive at the following formula.

( )( )

12 1

1

f xx x

f x= −

′

This point is also shown on the graph above and we can see from this graph that if we continue following this process will get a sequence of numbers that are getting very close the actual solution. This process is called Newton’s Method. Here is the general Newton’s Method Newton’s Method If xn is an approximation a solution of ( ) 0f x = and if ( ) 0nf x′ ≠ the next approximation is

given by,

( )( )1

nn n

n

f xx x

f x+ = −′

This should lead to the question of when do we stop? How many times do we go through this process? One of the more common stopping points in the process is to continue until two successive approximations agree to a given number of decimal places.


Calculus I


Before working any examples we should address two issues. First, we really do need to be solving ( ) 0f x = in order for Newton’s Method to be applied. This isn’t really all that much of

an issue but we do need to make sure that the equation is in this form prior to using the method. Secondly, we do need to somehow get our hands on an initial approximation to the solution (i.e. we need 0x somehow). One of the more common ways of getting our hands on 0x is to sketch

the graph of the function and use that to get an estimate of the solution which we then use as 0x . Another common method is if we know that there is a solution to a function in an interval then we can use the midpoint of the interval as 0x . Let’s work an example of Newton’s Method. Example 1 Use Newton’s Method to determine an approximation to the solution to cos x x= that lies in the interval [0,2]. Find the approximation to six decimal places. Solution First note that we weren’t given an initial guess. We were however, given an interval in which to look. We will use this to get our initial guess. As noted above the general rule of thumb in these cases is to take the initial approximation to be the midpoint of the interval. So, we’ll use 0 1x = as our initial guess. Next, recall that we must have the function in the form ( ) 0f x = . Therefore, we first rewrite

the equation as, cos 0x x− = We can now write down the general formula for Newton’s Method. Doing this will often simplify up the work a little so it’s generally not a bad idea to do this.

1cossin 1n n

x xx xx+

−= −

− −

Let’s now get the first approximation.

( )( )1

cos 1 11 0.7503638679

sin 1 1x

−= − =

− −

At this point we should point out that the phrase “six decimal places” does not mean just get x1 to six decimal places and then stop. Instead it means that we continue until two successive approximations agree to six decimal places. Given that stopping condition we clearly need to go at least one step farther.


Calculus I


( )( )2

cos 0.7503638679 0.75036386790.7503638679 0.7391128909

sin 0.7503638679 1x

−= − =

− −

Alright, we’re making progress. We’ve got the approximation to 1 decimal place. Let’s do another one, leaving the details of the computation to you. 3 0.7390851334x = We’ve got it to three decimal places. We’ll need another one. 4 0.7390851332x = And now we’ve got two approximations that agree to 9 decimal places and so we can stop. We will assume that the solution is approximately 4 0.7390851332x = .

In this last example we saw that we didn’t have to do too many computations in order for Newton’s Method to give us an approximation in the desired range of accuracy. This will not always be the case. Sometimes it will take many iterations through the process to get to the desired accuracy and on occasion it can fail completely. The following example is a little silly but it makes the point about the method failing.

Example 2 Use 0 1x = to find the approximation to the solution to 3 0x = . Solution Yes, it’s a silly example. Clearly the solution is 0x = , but it does make a very important point. Let’s get the general formula for Newton’s method.

13

1 23

3 213

nn n n n n

n

xx x x x xx

+−

= − = − = −

In fact we don’t really need to do any computations here. These computations get farther and farther away from the solution, 0x = ,with each iteration. Here are a couple of computations to make the point.

1

2

3

4

24

816

.

xxxxetc

= −== −=

So, in this case the method fails and fails spectacularly.


Calculus I


So, we need to be a little careful with Newton’s method. It will usually quickly find an approximation to an equation. However, there are times when it will take a lot of work or when it won’t work at all.


Calculus I


Business Applications In the final section of this chapter let’s take a look at some applications of derivatives in the business world. For the most part these are really applications that we’ve already looked at, but they are now going to be approached with an eye towards the business world. Let’s start things out with a couple of optimization problems. We’ve already looked at more than a few of these in previous sections so there really isn’t anything all that new here except for the fact that they are coming out of the business world. Example 3 An apartment complex has 250 apartments to rent. If they rent x apartments then their monthly profit, in dollars, is given by,

( ) 28 3200 80,000P x x x= − + −

How many apartments should they rent in order to maximize their profit? Solution All that we’re really being asked to do here is to maximize the profit subject to the constraint that x must be in the range 0 250x≤ ≤ . First, we’ll need the derivative and the critical point(s) that fall in the range 0 250x≤ ≤ . ( ) 3200

1616 3200 3200 16 0 200P x x x x′ = − + ⇒ − = ⇒ = =

Since the profit function is continuous and we have an interval with finite bounds we can find the maximum value by simply plugging in the only critical point that we have (which nicely enough in the range of acceptable answers) and the end points of the range. ( ) ( ) ( )0 80,000 200 240,000 250 220,000P P P= − = =

So, it looks like they will generate the most profit if they only rent out 200 of the apartments instead of all 250 of them. Note that with these problems you shouldn’t just assume that renting all the apartments will generate the most profit. Do not forget that there are all sorts of maintenance costs and that the more tenants renting apartments the more the maintenance costs will be. With this analysis we can see that, for this complex at least, something probably needs to be done to get the maximum profit more towards full capacity. This kind of analysis can help them determine just what they need to do to move towards that goal whether it be raising rent or finding a way to reduce maintenance costs. Note as well that because most apartment complexes have at least a few units empty after a tenant moves out and the like that it’s possible that they would actually like the maximum profit to fall


Calculus I


slightly under full capacity to take this into account. Again, another reason to not just assume that maximum profit will always be at the upper limit of the range. Let’s take a quick look at another problem along these lines. Example 4 A production facility is capable of producing 60,000 widgets in a day and the total daily cost of producing x widgets in a day is given by,

( ) 200,000,000250,000 0.08C x xx

= + +

How many widgets per day should they produce in order to minimize production costs? Solution Here we need to minimize the cost subject to the constraint that x must be in the range 0 60,000x≤ ≤ . Note that in this case the cost function is not continuous at the left endpoint and so we won’t be able to just plug critical points and endpoints into the cost function to find the minimum value. Let’s get the first couple of derivatives of the cost function.

( ) ( )2 3

200,000,000 400,000,0000.08C x C xx x

′ ′′= − =

The critical points of the cost function are,

2

2

2

200,000,0000.08 0

0.08 200,000,000

2,500,000,000 2,500,000,000 50,000

xx

x x

− =

=

= ⇒ = ± = ±

Now, clearly the negative value doesn’t make any sense in this setting and so we have a single critical point in the range of possible solutions : 50,000. Now, as long as 0x > the second derivative is positive and so, in the range of possible solutions the function is always concave up and so producing 50,000 widgets will yield the absolute minimum production cost. Now, we shouldn’t walk out of the previous two examples with the idea that the only applications to business are just applications we’ve already looked at but with a business “twist” to them. There are some very real applications to calculus that are in the business world and at some level that is the point of this section. Note that to really learn these applications and all of their intricacies you’ll need to take a business course or two or three. In this section we’re just going to scratch the surface and get a feel for some of the actual applications of calculus from the business world and some of the main “buzz” words in the applications.


Calculus I


Let’s start off by looking at the following example. Example 5 The production costs per week for producing x widgets is given by,

( ) 2500 350 0.09 , 0 1000C x x x x= + − ≤ ≤

Answer each of the following questions.

(a) What is the cost to produce the 301st widget? (b) What is the rate of change of the cost at 300x = ?

Solution (a) We can’t just compute ( )301C as that is the cost of producing 301 widgets while we are

looking for the actual cost of producing the 301st widget. In other words, what we’re looking for here is, ( ) ( )301 300 97,695.91 97, 400.00 295.91C C− = − =

So, the cost of producing the 301st widget is $295.91. (b) In this part all we need to do is get the derivative and then compute ( )300C′ .

( ) ( )350 0.18 300 296.00C x x C′ ′= − ⇒ =

Okay, so just what did we learn in this example? The cost to produce an additional item is called the marginal cost and as we’ve seen in the above example the marginal cost is approximated by the rate of change of the cost function, ( )C x . So, we define the marginal cost function to be

the derivative of the cost function or, ( )C x′ . Let’s work a quick example of this.

Example 6 The production costs per day for some widget is given by, ( ) 2 32500 10 0.01 0.0002C x x x x= − − +

What is the marginal cost when 200x = , 300x = and 400x = ? Solution So, we need the derivative and then we’ll need to compute some values of the derivative.

( )

( ) ( ) ( )

210 0.02 0.0006

200 10 300 38 400 78

C x x x

C C C

′ = − − +

′ ′ ′= = =

So, in order to produce the 201st widget it will cost approximately $10. To produce the 301st widget will cost around $38. Finally, to product the 401st widget it will cost approximately $78.


Calculus I


Note that it is important to note that ( )C n′ is the approximate cost of producing the ( )st1n +

item and NOT the nth item as it may seem to imply! Let’s now turn our attention to the average cost function. If ( )C x is the cost function for some

item then the average cost function is,

( ) ( )C xC x

x=

Here is the sketch of the average cost function from Example 4 above.

We can see from this that the average cost function has an absolute minimum. We can also see that this absolute minimum will occur at a critical point with ( ) 0C x′ = since it clearly will have

a horizontal tangent there. Now, we could get the average cost function, differentiate that and then find the critical point. However, this average cost function is fairly typical for average cost functions so let’s instead differentiate the general formula above using the quotient rule and see what we have.

( ) ( ) ( )2

x C x C xC x

x′ −

′ =

Now, as we noted above the absolute minimum will occur when ( ) 0C x′ = and this will in turn

occur when,

( ) ( ) ( ) ( ) ( )0C x

xC x C x C x C xx

′ ′− = ⇒ = =

So, we can see that it looks like for a typical average cost function we will get the minimum average cost when the marginal cost is equal to the average cost.


Calculus I


We should note however that not all average cost functions will look like this and so you shouldn’t assume that this will always be the case. Let’s now move onto the revenue and profit functions. First, let’s suppose that the price that some item can be sold at if there is a demand for x units is given by ( )p x . This function is

typically called either the demand function or the price function. The revenue function is then how much money is made by selling x items and is, ( ) ( )R x x p x= The profit function is then, ( ) ( ) ( ) ( ) ( )P x R x C x x p x C x= − = − Be careful to not confuse the demand function, ( )p x - lower case p, and the profit function,

( )P x - upper case P. Bad notation maybe, but there it is.

Finally, the marginal revenue function is ( )R x′ and the marginal profit function is ( )P x′

and these represent the revenue and profit respectively if one more unit is sold. Let’s take a quick look at an example of using these. Example 7 The weekly cost to produce x widgets is given by ( ) 2 375,000 100 0.03 0.000004 0 10000C x x x x x= + − + ≤ ≤

and the demand function for the widgets is given by, ( ) 200 0.005 0 10000p x x x= − ≤ ≤

Determine the marginal cost, marginal revenue and marginal profit when 2500 widgets are sold and when 7500 widgets are sold. Assume that the company sells exactly what they produce. Solution Okay, the first thing we need to do is get all the various functions that we’ll need. Here are the revenue and profit functions.

( ) ( )( ) ( )

2

2 2 3

2 3

200 0.005 200 0.005

200 0.005 75,000 100 0.03 0.000004

75,000 100 0.025 0.000004

R x x x x x

P x x x x x x

x x x

= − = −

= − − + − +

= − + + −

Now, all the marginal functions are,


Calculus I


( )( )( )

2

2

100 0.06 0.000012

200 0.01

100 0.05 0.000012

C x x x

R x x

P x x x

′ = − +

′ = −

′ = + −

The marginal functions when 2500 widgets are sold are, ( ) ( ) ( )2500 25 2500 175 2500 150C R P′ ′ ′= = =

The marginal functions when 7500 are sold are, ( ) ( ) ( )7500 325 7500 125 7500 200C R P′ ′ ′= = = −

So, upon producing and selling the 2501st widget it will cost the company approximately $25 to produce the widget and they will see an added $175 in revenue and $150 in profit. On the other hand when they produce and sell the 7501st widget it will cost an additional $325 and they will receive an extra $125 in revenue, but lose $200 in profit. We’ll close this section out with a brief discussion on maximizing the profit. If we assume that the maximum profit will occur at a critical point such that ( ) 0P x′ = we can then say the

following,

( ) ( ) ( ) ( ) ( )0P x R x C x R x C x′ ′ ′ ′ ′= − = ⇒ =

We then will know that this will be a maximum we also were to know that the profit was always concave down or, ( ) ( ) ( ) ( ) ( )0P x R x C x R x C x′′ ′′ ′′ ′′ ′′= − < ⇒ < So, if we know that ( ) ( )R x C x′′ ′′< then we will maximize the profit if ( ) ( )R x C x′ ′= or if the

marginal cost equals the marginal revenue. In this section we took a brief look at some of the ideas in the business world that involve calculus. Again, it needs to be stressed however that there is a lot more going on here and to really see how these applications are done you should really take some business courses. The point of this section was to just give a few ideas on how calculus is used in a field other than the sciences.


Calculus I



CALCULUS I Practice Problems


Paul Dawkins

Calculus I


Table of Contents Preface ............................................................................................................................................. 2 Applications of Derivatives ........................................................................................................... 2

Introduction .............................................................................................................................................. 3 Rates of Change ......................................................................................................................................... 3 Critical Points ............................................................................................................................................ 4 Minimum and Maximum Values .............................................................................................................. 5 Finding Absolute Extrema........................................................................................................................ 8 The Shape of a Graph, Part I ..................................................................................................................... 9 The Shape of a Graph, Part II ................................................................................................................. 11 The Mean Value Theorem ...................................................................................................................... 13 Optimization ........................................................................................................................................... 14 More Optimization Problems................................................................................................................. 14 Indeterminate Forms and L’Hospital’s Rule ......................................................................................... 15 Linear Approximations .......................................................................................................................... 16 Differentials............................................................................................................................................. 17 Newton’s Method .................................................................................................................................... 17 Business Applications ............................................................................................................................ 18


Calculus I


Preface Here are a set of practice problems for my Calculus I notes. If you are viewing the pdf version of this document (as opposed to viewing it on the web) this document contains only the problems themselves and no solutions are included in this document. Solutions can be found in a number of places on the site.




Note that some sections will have more problems than others and some will have more or less of a variety of problems. Most sections should have a range of difficulty levels in the problems although this will vary from section to section. I add problems to this document as I get a chance and so that means that sometimes some sections are apparently given more attention than others. Eventually I’ll have a full complement of problems for all the sections, however it takes time to write the problems, write the solutions for each problem and format them for presentation on the web. If there aren’t practice problems for a given section (or there aren’t that many problems in a section) then know that I will eventually get around to writing problems for the section (or more problems for the section as the case may be). Please do not email me asking when I’m going to get problems for a particular section written. As I’ve already said, I write problems as I have time to do so. Please understand that I have my own classes to teach and other responsibilities to my department that I have to also take care of. This web site and all of its information has been written and maintained in my spare time and that often means that I don’t have as much time to work on it as I’d like to.




Calculus I


Introduction Here are a set of practice problems for the Derivatives chapter of my Calculus I notes. If you are viewing the pdf version of this document (as opposed to viewing it on the web) this document contains only the problems themselves and no solutions are included in this document. Solutions can be found in a number of places on the site.




Note that some sections will have more problems than others and some will have more or less of a variety of problems. Most sections should have a range of difficulty levels in the problems although this will vary from section to section. Here is a list of topics in this chapter that have practice problems written for them. Rates of Change Critical Points Minimum and Maximum Values Finding Absolute Extrema The Shape of a Graph, Part I The Shape of a Graph, Part II The Mean Value Theorem Optimization Problems More Optimization Problems L’Hospital’s Rule and Indeterminate Forms Linear Approximations Differentials Newton’s Method Business Applications

Rates of Change As noted in the text for this section the purpose of this section is only to remind you of certain types of applications that were discussed in the previous chapter. As such there aren’t any problems written for this section. Instead here is a list of links (note that these will only be active



Calculus I


links in the web version and not the pdf version) to problems from the relevant sections from the previous chapter. Each of the following sections has a selection of increasing/decreasing problems towards the bottom of the problem set.

Differentiation Formulas Product & Quotient Rules Derivatives of Trig Functions Derivatives of Exponential and Logarithm Functions Chain Rule

Related Rates problems are in the Related Rates section.

Critical Points Determine the critical points of each of the following functions. 1. ( ) 3 28 81 42 8f x x x x= + − −

2. ( ) 3 4 51 80 5 2R t t t t= + + −

3. ( ) 3 22 7 3 2g w w w w= − − −

4. ( ) 6 5 42 8g x x x x= − +

5. ( ) 3 24 3 9 12h z z z z= − + +

6. ( ) ( ) ( )34 22 8 9Q x x x= − −

7. ( ) 2

42 8

zf zz z

+=

+ +

8. ( ) 2

12 15

xR xx x

−=

+ −

9. ( ) 25 6r y y y= −


Calculus I


10. ( ) ( )1

2 315 3 8 7h t t t t = − − − +

11. ( ) ( )4coss z z z= −

12. ( ) ( )3 92sin y yf y = +

13. ( ) ( )2sin 3 1V t t= +

14. ( ) 9 25 xf x x −= e

15. ( ) 3 22 7w w wg w − −= e

16. ( ) ( )2ln 4 14R x x x= + +

17. ( ) ( )3 7ln 8 2A t t t= − +

Minimum and Maximum Values 1. Below is the graph of some function, ( )f x . Identify all of the relative extrema and absolute

extrema of the function.

2. Below is the graph of some function, ( )f x . Identify all of the relative extrema and absolute



Calculus I


3. Sketch the graph of ( ) 2 4g x x x= − and identify all the relative extrema and absolute extrema

of the function on each of the following intervals. (a) ( ),−∞ ∞

(b) [ ]1,4−

(c) [ ]1,3

(d) [ ]3,5

(e) ( ]1,5−

4. Sketch the graph of ( ) ( )34h x x= − + and identify all the relative extrema and absolute

extrema of the function on each of the following intervals. (a) ( ),−∞ ∞

(b) [ ]5.5, 2− −

(c) [ )4, 3− − (d) [ ]4, 3− −

5. Sketch the graph of some function on the interval [ ]1,6 that has an absolute maximum at

6x = and an absolute minimum at 3x = . 6. Sketch the graph of some function on the interval [ ]4,3− that has an absolute maximum at

3x = − and an absolute minimum at 2x = . 7. Sketch the graph of some function that meets the following conditions :

(a) The function is continuous. (b) Has two relative minimums.


Calculus I


(b) One of relative minimums is also an absolute minimum and the other relative minimum is not an absolute minimum.

(c) Has one relative maximum. (d) Has no absolute maximum.


Calculus I


Finding Absolute Extrema For each of the following problems determine the absolute extrema of the given function on the specified interval. 1. ( ) 3 28 81 42 8f x x x x= + − − on [ ]8,2−

2. ( ) 3 28 81 42 8f x x x x= + − − on [ ]4,2−

3. ( ) 3 4 51 80 5 2R t t t t= + + − on [ ]4.5, 4−

4. ( ) 3 4 51 80 5 2R t t t t= + + − on [ ]0,7

5. ( ) 3 24 3 9 12h z z z z= − + + on [ ]2,1−

6. ( ) 4 3 23 26 60 11g x x x x= − + − on [ ]1,5

7. ( ) ( ) ( )34 22 8 9Q x x x= − − on [ ]3,3−

8. ( ) ( )532 2h w w w= + on 5 12 2,−

9. ( ) 2

42 8

zf zz z

+=

+ + on [ ]10,0−

10. ( ) ( )2

2 310A t t t= − on [ ]2, 10.5

11. ( ) ( )3 92sin y yf y = + on [ ]10,15−

12. ( ) 3 22 7w w wg w − −= e on 512 2,−

13. ( ) ( )2ln 4 14R x x x= + + on [ ]4,2−


Calculus I


The Shape of a Graph, Part I For problems 1 & 2 the graph of a function is given. Determine the open intervals on which the function increases and decreases. 1.

2.

3. Below is the graph of the derivative of a function. From this graph determine the open intervals in which the function increases and decreases.


Calculus I


4. This problem is about some function. All we know about the function is that it exists everywhere and we also know the information given below about the derivative of the function. Answer each of the following questions about this function.

(a) Identify the critical points of the function. (b) Determine the open intervals on which the function increases and decreases. (c) Classify the critical points as relative maximums, relative minimums or neither.

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )5 0 2 0 4 0 8 0

0 on 5, 2 , 2,4 , 8, 0 on , 5 , 4,8

f f f f

f x f x

′ ′ ′ ′− = − = = =

′ ′< − − − ∞ > −∞ −

For problems 5 – 12 answer each of the following.


5. ( ) 3 22 9 60f x x x x= − −

6. ( ) 3 4 550 40 5 4h t t t t= + − −

7. 3 22 10 12 12y x x x= − + − 8. ( ) ( )cos 3 2p x x x= + on 3

2 , 2−

9. ( ) ( )22 5 14sin zR z z= − − on [ ]10,7−

10. ( ) 2 3 7h t t t= −


Calculus I


11. ( )21

22 wf w w −= e

12. ( ) ( )22ln 1g x x x= − +

13. For some function, ( )f x , it is known that there is a relative maximum at 4x = . Answer

each of the following questions about this function. (a) What is the simplest form for the derivative of this function? Note : There really are many

possible forms of the derivative so to make the rest of this problem as simple as possible you will want to use the simplest form of the derivative that you can come up with.

(b) Using your answer from (a) determine the most general form of the function. (c) Given that ( )4 1f = find a function that will have a relative maximum at 4x = . Note :

You should be able to use your answer from (b) to determine an answer to this part. 14. Given that ( )f x and ( )g x are increasing functions. If we define ( ) ( ) ( )h x f x g x= +

show that ( )h x is an increasing function.

15. Given that ( )f x is an increasing function and define ( ) ( ) 2h x f x= . Will ( )h x be an

increasing function? If yes, prove that ( )h x is an increasing function. If not, can you determine

any other conditions needed on the function ( )f x that will guarantee that ( )h x will also

increase?

The Shape of a Graph, Part II 1. The graph of a function is given below. Determine the open intervals on which the function is concave up and concave down.


Calculus I


2. Below is the graph the 2nd derivative of a function. From this graph determine the open intervals in which the function is concave up and concave down.


(a) Determine a list of possible inflection points for the function. (b) Determine the open intervals on which the function is concave up and concave down. (c) Determine the inflection points of the function.

3. ( ) 2 312 6f x x x= + −

4. ( ) 4 312 84 4g z z z z= − + +

5. ( ) 4 3 212 6 36 2h t t t t t= + + − +

6. ( ) ( )28 5 2 cos 3h w w w w= − + − on [ ]1,2−

7. ( ) ( )234R z z z= +

8. ( ) 24 xh x −= e


(a) Identify the critical points of the function. (b) Determine the open intervals on which the function increases and decreases. (c) Classify the critical points as relative maximums, relative minimums or neither. (d) Determine the open intervals on which the function is concave up and concave down.


Calculus I


(e) Determine the inflection points of the function. (f) Use the information from steps (a) – (e) to sketch the graph of the function.

9. ( ) 5 45 8g t t t= − +

10. ( ) 3 45 8f x x x= − −

11. ( ) 4 3 22 12h z z z z= − −

12. ( ) ( )23 8sin tQ t t= − on [ ]7,4−

13. ( ) ( )43 2f x x x= −

14. ( ) 4wP w w= e on [ ]1

42,−

15. Determine the minimum degree of a polynomial that has exactly one inflection point. 16. Suppose that we know that ( )f x is a polynomial with critical points 1x = − , 2x = and

6x = . If we also know that the 2nd derivative is ( ) 23 14 4f x x x′′ = + − . If possible, classify

each of the critical points as relative minimums, relative maximums. If it is not possible to classify the critical points clearly explain why they cannot be classified.

The Mean Value Theorem For problems 1 & 2 determine all the number(s) c which satisfy the conclusion of Rolle’s Theorem for the given function and interval. 1. ( ) 2 2 8f x x x= − − on [ ]1,3−

2. ( ) 2 32g t t t t= − − on [ ]2,1−

For problems 3 & 4 determine all the number(s) c which satisfy the conclusion of the Mean Value Theorem for the given function and interval. 3. ( ) 3 24 8 7 2h z z z z= − + − on [ ]2,5


Calculus I


4. ( ) 38 tA t t −= + e on [ ]2,3−

5. Suppose we know that ( )f x is continuous and differentiable on the interval [ ]7,0− , that

( )7 3f − = − and that ( ) 2f x′ ≤ . What is the largest possible value for ( )0f ?

6. Show that ( ) 3 27 25 8f x x x x= − + + has exactly one real root.

Optimization 1. Find two positive numbers whose sum is 300 and whose product is a maximum. 2. Find two positive numbers whose product is 750 and for which the sum of one and 10 times the other is a minimum. 3. Let x and y be two positive numbers such that 2 50x y+ = and ( ) ( )1 2x y+ + is a maximum.

4. We are going to fence in a rectangular field. If we look at the field from above the cost of the vertical sides are $10/ft, the cost of the bottom is $2/ft and the cost of the top is $7/ft. If we have $700 determine the dimensions of the field that will maximize the enclosed area. 5. We have 45 m2 of material to build a box with a square base and no top. Determine the dimensions of the box that will maximize the enclosed volume. 6. We want to build a box whose base length is 6 times the base width and the box will enclose 20 in3. The cost of the material of the sides is $3/in2 and the cost of the top and bottom is $15/in2. Determine the dimensions of the box that will minimize the cost. 7. We want to construct a cylindrical can with a bottom but no top that will have a volume of 30 cm3. Determine the dimensions of the can that will minimize the amount of material needed to construct the can. 8. We have a piece of cardboard that is 50 cm by 20 cm and we are going to cut out the corners and fold up the sides to form a box. Determine the height of the box that will give a maximum volume.

More Optimization Problems


Calculus I


1. We want to construct a window whose middle is a rectangle and the top and bottom of the window are semi-circles. If we have 50 meters of framing material what are the dimensions of the window that will let in the most light? 2. Determine the area of the largest rectangle that can be inscribed in a circle of radius 1. 3. Find the point(s) on 23 2x y= − that are closest to ( )4,0− .

4. An 80 cm piece of wire is cut into two pieces. One piece is bent into an equilateral triangle and the other will be bent into a rectangle with one side 4 times the length of the other side. Determine where, if anywhere, the wire should be cut to maximize the area enclosed by the two figures. 5. A line through the point ( )2,5 forms a right triangle with the x-axis and y-axis in the 1st

quadrant. Determine the equation of the line that will minimize the area of this triangle. 6. A piece of pipe is being carried down a hallway that is 18 feet wide. At the end of the hallway there is a right-angled turn and the hallway narrows down to 12 feet wide. What is the longest pipe (always keeping it horizontal) that can be carried around the turn in the hallway? 7. Two 10 meter tall poles are 30 meters apart. A length of wire is attached to the top of each pole and it is staked to the ground somewhere between the two poles. Where should the wire be staked so that the minimum amount of wire is used?

Indeterminate Forms and L’Hospital’s Rule Use L’Hospital’s Rule to evaluate each of the following limits.

1. 3 2

22

7 10lim6x

x x xx x→

− ++ −

2. ( )

24

sinlim

16w

ww

π→− −

3. ( )

2

ln 3limt

tt→∞

4. ( )

( )

2

220

sin 2 7 2lim

1z

z z zz z→

+ −

+


Calculus I


5. 2

1limx x

x−→−∞ e

6. 2 4

lim2z

z

zz

z→∞

+−

ee

7. 3lim ln 1

tt

t→∞

+

8. ( )2 2

0lim ln 4w

w w+→

9. ( ) ( )21lim 1 tanx

x xπ+→

−

10. ( ) 21

0lim cos 2y

yy+→

11. 1

limx

x xx→∞

+ e

Linear Approximations For problems 1 & 2 find a linear approximation to the function at the given point. 1. ( ) 2 103 xf x x −= e at 5x =

2. ( ) 4 36 3 7h t t t t= − + − at 3t = −

3. Find the linear approximation to ( ) 4g z z= at 2z = . Use the linear approximation to

approximate the value of 4 3 and 4 10 . Compare the approximated values to the exact values. 4. Find the linear approximation to ( ) ( )cos 2f t t= at 1

2t = . Use the linear approximation to

approximate the value of ( )cos 1 and ( )cos 9 . Compare the approximated values to the exact

values. 5. Without using any kind of computational aid use a linear approximation to estimate the value of 0.1e .


Calculus I


Differentials For problems 1 – 3 compute the differential of the given function. 1. ( ) ( )2 secf x x x= −

2. 4 2 4x x xw − += e

3. ( ) ( ) ( )ln 2 sin 2h z z z=

4. Compute dy and y∆ for 2xy = e as x changes from 3 to 3.01.

5. Compute dy and y∆ for 5 32 7y x x x= − + as x changes from 6 to 5.9. 6. The sides of a cube are found to be 6 feet in length with a possible error of no more than 1.5 inches. What is the maximum possible error in the volume of the cube if we use this value of the length of the side to compute the volume?

Newton’s Method

For problems 1 & 2 use Newton’s Method to determine 2x for the given function and given value

of 0x . 1. ( ) 3 27 8 3f x x x x= − + − , 0 5x =

2. ( ) ( ) 2cosf x x x x= − , 0 1x =

For problems 3 & 4 use Newton’s Method to find the root of the given equation, accurate to six decimal places, that lies in the given interval. 3. 4 35 9 3 0x x x− + + = in [ ]4,6

4. 22 5 xx + = e in [ ]3, 4


Calculus I


For problems 5 & 6 use Newton’s Method to find all the roots of the given equation accurate to six decimal places. 5. 3 2 15 1 0x x x− − + = 6. ( )22 sinx x− =

Business Applications 1. A company can produce a maximum of 1500 widgets in a year. If they sell x widgets during the year then their profit, in dollars, is given by, ( ) 2 31

330,000,000 360,000 750P x x x x= − + − How many widgets should they try to sell in order to maximize their profit? 2. A management company is going to build a new apartment complex. They know that if the complex contains x apartments the maintenance costs for the building, landscaping etc. will be, ( ) 24000 14 0.04C x x x= + − The land they have purchased can hold a complex of at most 500 apartments. How many apartments should the complex have in order to minimize the maintenance costs? 3. The production costs, in dollars, per day of producing x widgets is given by, ( ) 2 31750 6 0.04 0.0003C x x x x= + − + What is the marginal cost when 175x = and 300x = ? What do your answers tell you about the production costs? 4. The production costs, in dollars, per month of producing x widgets is given by,

( ) 10000200 0.5C x xx

= + +

What is the marginal cost when 200x = and 500x = ? What do your answers tell you about the production costs? 5. The production costs, in dollars, per week of producing x widgets is given by,


Calculus I


( ) 2 34000 32 0.08 0.00006C x x x x= − + + and the demand function for the widgets is given by, ( ) 2250 0.02 0.001p x x x= + − What is the marginal cost, marginal revenue and marginal profit when 200x = and 400x = ? What do these numbers tell you about the cost, revenue and profit?


CALCULUS I Solutions to Practice Problems


Paul Dawkins

Calculus I



Rates of Change ......................................................................................................................................... 2 Critical Points ............................................................................................................................................ 2 Minimum and Maximum Values ............................................................................................................ 16 Finding Absolute Extrema...................................................................................................................... 27 The Shape of a Graph, Part I ................................................................................................................... 43 The Shape of a Graph, Part II ................................................................................................................. 66 The Mean Value Theorem ...................................................................................................................... 94 Optimization ........................................................................................................................................... 99 More Optimization Problems............................................................................................................... 112 Indeterminate Forms and L’Hospital’s Rule ....................................................................................... 127 Linear Approximations ........................................................................................................................ 141 Differentials........................................................................................................................................... 145 Newton’s Method .................................................................................................................................. 148 Business Applications .......................................................................................................................... 159

Preface Here are the solutions to the practice problems for my Calculus I notes. Some solutions will have more or less detail than other solutions. The level of detail in each solution will depend up on several issues. If the section is a review section, this mostly applies to problems in the first chapter, there will probably not be as much detail to the solutions given that the problems really should be review. As the difficulty level of the problems increases less detail will go into the basics of the solution under the assumption that if you’ve reached the level of working the harder problems then you will probably already understand the basics fairly well and won’t need all the explanation. This document was written with presentation on the web in mind. On the web most solutions are broken down into steps and many of the steps have hints. Each hint on the web is given as a popup however in this document they are listed prior to each step. Also, on the web each step can be viewed individually by clicking on links while in this document they are all showing. Also, there are liable to be some formatting parts in this document intended for help in generating the web pages that haven’t been removed here. These issues may make the solutions a little difficult to follow at times, but they should still be readable.



Calculus I


Rates of Change As noted in the text for this section the purpose of this section is only to remind you of certain types of applications that were discussed in the previous chapter. As such there aren’t any problems written for this section. Instead here is a list of links (note that these will only be active links in the web version and not the pdf version) to problems from the relevant sections from the previous chapter. Each of the following sections has a selection of increasing/decreasing problems towards the bottom of the problem set.



Critical Points 1. Determine the critical points of ( ) 3 28 81 42 8f x x x x= + − − .

Step 1 We’ll need the first derivative to get the answer to this problem so let’s get that. ( ) ( ) ( )224 162 42 6 7 4 1f x x x x x′ = + − = + − Factoring the derivative as much as possible will help with the next step. Step 2 Recall that critical points are simply where the derivative is zero and/or doesn’t exist. In this case the derivative is just a polynomial and we know that exists everywhere and so we don’t need to worry about that. So, all we need to do is set the derivative equal to zero and solve for the critical points. ( )( ) 1

46 7 4 1 0 7,x x x x+ − = ⇒ = − = 2. Determine the critical points of ( ) 3 4 51 80 5 2R t t t t= + + − .


Calculus I


Step 1 We’ll need the first derivative to get the answer to this problem so let’s get that. ( ) ( )( )2 3 4 2240 20 10 10 4 6R t t t t t t t′ = + − = − + − Factoring the derivative as much as possible will help with the next step. Step 2 Recall that critical points are simply where the derivative is zero and/or doesn’t exist. In this case the derivative is just a polynomial and we know that exists everywhere and so we don’t need to worry about that. So, all we need to do is set the derivative equal to zero and solve for the critical points. ( )( )210 4 6 0 0, 4, 6t t t t t t− + − = ⇒ = = − = 3. Determine the critical points of ( ) 3 22 7 3 2g w w w w= − − − .

Step 1 We’ll need the first derivative to get the answer to this problem so let’s get that. ( ) 26 14 3g w w w′ = − − Step 2 Recall that critical points are simply where the derivative is zero and/or doesn’t exist. In this case the derivative is just a polynomial and we know that exists everywhere and so we don’t need to worry about that. So, all we need to do is set the derivative equal to zero and solve for the critical points.

2 14 268 7 676 14 3 012 6

w w w ± ±− − = ⇒ = =

As we can see in this case we needed to use the quadratic formula to find the critical points. Not all quadratics will factor so don’t forget about the quadratic formula! 4. Determine the critical points of ( ) 6 5 42 8g x x x x= − + .

Step 1 We’ll need the first derivative to get the answer to this problem so let’s get that.


Calculus I


( ) ( )5 4 3 3 26 10 32 2 3 5 16g x x x x x x x′ = − + = − + Factoring the derivative as much as possible will help with the next step. Step 2 Recall that critical points are simply where the derivative is zero and/or doesn’t exist. In this case the derivative is just a polynomial and we know that exists everywhere and so we don’t need to worry about that. So, all we need to do is set the derivative equal to zero and solve for the critical points. ( )3 2 3 22 3 5 16 0 2 0 OR 3 5 16 0x x x x x x− + = ⇒ = − + =

From the first term we clearly see that 0x = is a critical point. The second term does not factor and we we’ll need to use the quadratic formula to solve this equation.

5 167 5 167

6 6ix ± − ±

= =

Remember that not all quadratics will factor so don’t forget about the quadratic formula! Step 3 Now, recall that we don’t use complex numbers in this class and so the solutions from the second term are not critical points. Therefore, the only critical point of this function is,

0x = 5. Determine the critical points of ( ) 3 24 3 9 12h z z z z= − + + .

Step 1 We’ll need the first derivative to get the answer to this problem so let’s get that. ( ) 212 6 9h z z z′ = − + Step 2 Recall that critical points are simply where the derivative is zero and/or doesn’t exist. In this case the derivative is just a polynomial and we know that exists everywhere and so we don’t need to worry about that. So, all we need to do is set the derivative equal to zero and solve for the critical points.


Calculus I


2 6 396 1 1112 6 9 024 4

iz z z ± − ±− + = ⇒ = =

As we can see in this case we needed to use the quadratic formula to solve the quadratic. Remember that not all quadratics will factor so don’t forget about the quadratic formula! Step 3 Now, recall that we don’t use complex numbers in this class and so the solutions are not critical points. Therefore, there are no critical points for this function. Do not get excited about there being no critical points for a function. There is no rule that says that every function has to have critical points!

6. Determine the critical points of ( ) ( ) ( )34 22 8 9Q x x x= − − .


( ) ( ) ( )( ) ( ) ( )( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

3 23 42 2

23 2 2

2 23 32 2 2 2

4 2 8 8 9 2 8 3 9 2

2 2 8 9 16 9 3 2 8

2 2 8 9 40 6 144 4 2 8 9 20 3 72

Q x x x x x x

x x x x x

x x x x x x x x

′ = − − − + − −

= − − − − + −

= − − − + + = − − − − −

Factoring the derivative as much as possible will help with the next step. For this problem (unlike some of the previous problems) this extra factoring is all but required to make this easier to finish. Step 2 Recall that critical points are simply where the derivative is zero and/or doesn’t exist. In this case the derivative is just a polynomial, (admittedly a somewhat messy polynomial) and we know that exists everywhere and so we don’t need to worry about that. So, all we need to do is set the derivative equal to zero and solve for the critical points. ( ) ( )23 2 24 2 8 9 20 3 72 0x x x x − − − − − = From this we get the following three equations that we need to solve.


Calculus I


( )

( )

3

22

2

2 8 0

9 0

20 3 72 0

x

x

x x

− =

− =

− − =

For the first two equations all we really need to do is set the quantity inside the parenthesis to zero (the exponent on the parenthesis won’t affect the solution) and the third requires the quadratic formula.

142 8 0x x− = ⇒ =

2 9 0 3x x− = ⇒ = ±

( ) ( )

( )

22 3 3 4 20 72 3 576920 3 72 0

2 20 40x x x

± − − ±− − = ⇒ = =

So, we get the 5 critical points boxed in above.

7. Determine the critical points of ( ) 2

42 8

zf zz z

+=

+ +.


( )( )( ) ( ) ( )

( ) ( )( )

( )

2 22

2 2 22 2 2

1 2 8 4 4 1 2 8 22 16 4

2 8 2 8 2 8

z z z z z zz zf zz z z z z z

+ + − + + − + −− − +′ = = =+ + + + + +

The “-2” was factored out of the numerator only to make it a little nicer for the next step and doesn’t really need to be done. Step 2 Recall that critical points are simply where the derivative is zero and/or doesn’t exist. In this case the derivative is a rational expression. Therefore we know that the derivative will be zero if the numerator is zero (and the denominator is also not zero for the same values of course). We also know that the derivative won’t exist if we get division by zero. So, all we need to do is set the numerator and denominator equal to zero and solve. Note as well that the “-2” we factored out of the numerator will not affect where it is zero and so can be ignored. Likewise, the exponent on the whole denominator will not affect where it is zero and so can also be ignored. This means we need to solve the following two equations.


Calculus I


2 8 728 2 0 4 3 22

z z z − ±+ − = ⇒ = = − ±

2 1 63 1 632 8 04 4

iz z z − ± − − ±+ + = ⇒ = =

As we can see in this case we needed to use the quadratic formula both of the quadratic equations. Remember that not all quadratics will factor so don’t forget about the quadratic formula! Step 3 Now, recall that we don’t use complex numbers in this class and so the solutions from where the denominator is zero (i.e. the derivative doesn’t exist) won’t be critical points. Therefore, the only critical points of this function are,

4 3 2x = ±

8. Determine the critical points of ( ) 2

12 15

xR xx x

−=

+ −.


( )( ) ( ) ( ) ( )

( ) ( )

2 2

2 22 2

1 2 15 1 2 2 2 13

2 15 2 15

x x x x x xR xx x x x

− + − − − + − +′ = =+ − + −

Step 2 Recall that critical points are simply where the derivative is zero and/or doesn’t exist. In this case the derivative is a rational expression. Therefore we know that the derivative will be zero if the numerator is zero (and the denominator is also not zero for the same values of course). We also know that the derivative won’t exist if we get division by zero. So, all we need to do is set the numerator and denominator equal to zero and solve. Note that the exponent on the whole denominator will not affect where it is zero and so can be ignored. This means we need to solve the following two equations.

2 2 482 13 0 1 2 32

x x x ±− + = ⇒ = = ±

( )( )2 2 15 5 3 0 5, 3x x x x x+ − = + − = ⇒ = −


Calculus I


As we can see in this case we needed to use the quadratic formula on the first quadratic equation. Remember that not all quadratics will factor so don’t forget about the quadratic formula! Step 3 Now, recall that we don’t use complex numbers in this class and so the solutions from where the numerator is zero won’t be critical points. Also recall that a point will only be a critical point if the function (not the derivative, but the original function) exists at the point. For this problem we found two values where the derivative doesn’t exist, however the function also doesn’t exist at these points and so neither of these will be critical points either. Therefore, this function has no critical points. Do not get excited about this when it happens. Not all functions will have critical points!

9. Determine the critical points of ( ) 25 6r y y y= − .


( ) ( )( )( )

421 5

5 42 5

2 62 6 65 6

yr y y y yy y

− −′ = − − =−

We took the term with the negative exponent to the denominator for the discussion in the next step. While it doesn’t really need to be done this will make sure that there are no inadvertent mistakes down the road. Step 2 Recall that critical points are simply where the derivative is zero and/or doesn’t exist. In this case the derivative is a rational expression. Therefore we know that the derivative will be zero if the numerator is zero (and the denominator is also not zero for the same values of course). We also know that the derivative won’t exist if we get division by zero. So, all we need to do is set the numerator and denominator equal to zero and solve. Note that the exponent on the whole denominator will not affect where it is zero and so can be ignored. This means we need to solve the following two equations. 2 6 0 3y y− = ⇒ =

( )2 6 6 0 0,6y y y y y− = − = ⇒ = Step 3


Calculus I


Note as well that the reason for moving the term to the denominator as we did in the first step is to make it clear that the last two critical points are critical points because the derivative does not exist at those points and not because the derivative is zero at those points. Also note that they are critical points because the function does exist at these points. Therefore, along with the first critical point (where the derivative is zero), we get the following critical points for this function.

0, 3, 6y =

10. Determine the critical points of ( ) ( )1

2 315 3 8 7h t t t t = − − − + .


( ) ( )( )( ) ( ) ( )

( ) ( ) ( )

( ) ( )

1 2 12 2 213 3 3

3 22 3

2 2

2 22 23 3

3 2 88 7 3 2 8 8 7 8 7

3 8 7

3 8 7 3 2 8 5 38 45

3 8 7 3 8 7

t th t t t t t t t t t

t t

t t t t t t

t t t t

− − −′ = − + − − − − + = − + −

− +

− + − − − − += =

− + − +

After differentiating we moved the term with the negative exponent to the denominator and then combined everything into a single term. This will help with the next step considerably. Step 2 Recall that critical points are simply where the derivative is zero and/or doesn’t exist. Because we moved the term with the negative exponent to the denominator and then combined everything into a single term we now have written the derivative as a rational expression. Therefore we know that the derivative will be zero if the numerator is zero (and the denominator is also not zero for the same values of course). We also know that the derivative won’t exist if we get division by zero. So, all we need to do is set the numerator and denominator equal to zero and solve. Note that the exponent on the whole denominator will not affect where it is zero and so can be ignored. This means we need to solve the following two equations.

2 38 544 19 2 345 38 45 010 5

t t t ± ±− + = ⇒ = =


Calculus I


( ) ( )2 8 7 7 1 0 1, 7t t t t t− + = − − = ⇒ = Step 3 Note that because we combined all the terms in the derivative into a single term it was much easier to determine the critical points for this function. If we had not combined the terms the solving work would have been more complicated, although not impossible. Doing this also makes it clear that the last two critical points are critical points because the derivative does not exist at those points and not because the derivative is zero at those points. Also note that they are critical points because the function does exist at these points. Therefore, along with the first two critical points (where the derivative is zero), we get the following critical points for this function.

19 2 341, 7,5

t ±=

11. Determine the critical points of ( ) ( )4coss z z z= − .

Step 1 We’ll need the first derivative to get the answer to this problem so let’s get that. ( ) ( )4sin 1s z z′ = − − Step 2 Recall that critical points are simply where the derivative is zero and/or doesn’t exist. This derivative exists everywhere and so we don’t need to worry about that. Therefore all we need to do is determine where the derivative is zero. So, all we need to do is solve the equation, ( ) ( ) ( )11 1

4 44sin 1 0 sin sin 0.2527z z z −− − = → = − → = − = −

This is the answer we got from a calculator and we could use this or we could use the equivalent positive angle : 2 0.2527 6.0305π − = . Either can be used, but we’ll use the positive one for this problem. Now, a quick look at a unit circle gives us a second solution of 0.2527 3.3943π + = . Finally all possible solutions to this equation, and hence, all the critical points of the original function are,


Calculus I


6.0305 20, 1, 2, ,

3.3943 2z n

nz n

ππ

= += ± ± ±

= +K

If you don’t remember how to solve trig equations you should go back and review those sections in the Review Chapter of the notes.

12. Determine the critical points of ( ) ( )3 92sin y yf y = + .


( ) ( )1 23 3 9cos yf y′ = +

Step 2 Recall that critical points are simply where the derivative is zero and/or doesn’t exist. This derivative exists everywhere and so we don’t need to worry about that. Therefore all we need to do is determine where the derivative is zero. So, all we need to do is solve the equation, ( ) ( ) ( )11 2 2 2

3 3 9 3 3 3 3cos 0 cos cos 2.3005y y y −+ = → = − → = − = This is the answer we got from a calculator and a quick look at a unit circle gives us a second solution of either -2.3005 or if you want the positive equivalent we could use 2 2.3005 3.9827π − = . For this problem we’ll use the positive one, although the negative one could just as easily be used if you wanted to.

All possible solutions to ( ) 23 3cos y = − are then,

3

3

2.3005 20, 1, 2, ,

3.9827 2

y

yn

nn

π

π

= += ± ± ±

= +K

Finally solving for y gives all the critical points of the function.

6.9015 60, 1, 2, ,

11.9481 6y n

ny n

ππ

= += ± ± ±

= +K

If you don’t remember how to solve trig equations you should go back and review those sections in the Review Chapter of the notes.


Calculus I


13. Determine the critical points of ( ) ( )2sin 3 1V t t= + .

Step 1 We’ll need the first derivative to get the answer to this problem so let’s get that. ( ) ( ) ( )6sin 3 cos 3V t t t′ = Step 2 Recall that critical points are simply where the derivative is zero and/or doesn’t exist. This derivative exists everywhere and so we don’t need to worry about that. Therefore all we need to do is determine where the derivative is zero. So, all we need to do is solve the equation, ( ) ( ) ( ) ( )6sin 3 cos 3 0 sin 3 0 or cos 3 0t t t t= → = =

Step 3 So, we now need to solve these two trig equations. From a quick look at a unit circle we can see that sine is zero at 0 and π and so all solutions to

( )sin 3 0t = are then, 231 23 3

3 0 20, 1, 2, ,

3 2t nt n

nt nt n

πππ ππ π

== +→ = ± ± ±

= += +K

Another look at a unit circle and we can see that cosine is zero at 2

π and 32π and so all solutions

to ( )cos 3 0t = are then, 2

6 323 2

2 32

3 20, 1, 2, ,

3 2t nt n

nt nt n

ππ

ππππππ

= += +→ = ± ± ±

= += +K

Therefore, critical points of the function are,

2 1 2 2 23 3 3 6 3 2 3, , , 0, 1, 2, ,t n t n t n t n nπ ππ π π π π= = + = + = + = ± ± ± K

If you don’t remember how to solve trig equations you should go back and review those sections in the Review Chapter of the notes. 14. Determine the critical points of ( ) 9 25 xf x x −= e .


Calculus I


Step 1 We’ll need the first derivative to get the answer to this problem so let’s get that. ( ) ( ) ( )9 2 9 2 9 25 5 2 5 1 2x x xf x x x− − −′ = + − = −e e e We did some quick factoring to help with the next step and while it doesn’t technically need to be done it will significantly reduce the amount work required in the next step. Step 2 Recall that critical points are simply where the derivative is zero and/or doesn’t exist. This derivative exists everywhere and so we don’t need to worry about that. Therefore all we need to do is determine where the derivative is zero. Notice as well that because we know that exponential functions are never zero and so the derivative will only be zero if, 1

21 2 0x x− = → =

So, we have a single critical point, 12x = , for this function.

15. Determine the critical points of ( ) 3 22 7w w wg w − −= e .


( ) ( ) ( ) ( )3 2 3 22 2 7 2 73 4 7 3 7 1w w w w w wg w w w w w− − − −′ = − − = − +e e We did some quick factoring to help with the next step. Step 2 Recall that critical points are simply where the derivative is zero and/or doesn’t exist. This derivative exists everywhere and so we don’t need to worry about that. Therefore all we need to do is determine where the derivative is zero. Notice as well that because we know that exponential functions are never zero and so the derivative will only be zero if,


Calculus I


( )( ) 733 7 1 0 , 1w w w− + = → = −

So, we have a two critical points, 73w = and 1w = − for this function.

16. Determine the critical points of ( ) ( )2ln 4 14R x x x= + + .


( ) 2

2 44 14xR x

x x+′ =

+ +

Step 2 Recall that critical points are simply where the derivative is zero and/or doesn’t exist. In this case the derivative is a rational expression. So, we know that the derivative will be zero if the numerator is zero (and the denominator is also not zero for the same values of course). We also know that the derivative won’t exist if we get division by zero. However, in this case note that the denominator is also the polynomial that is inside the logarithm and so any values of x for which the denominator is zero will not be in the domain of the original function (i.e. the function, ( )R x , won’t exist at those values of x because we can’t take the logarithm of zero).

Therefore, these points will not be critical points and we don’t need to bother determining where the derivative will be zero. So, setting the numerator equal to zero gives, 2 4 0 2x x+ = ⇒ = −

Step 3 As a final step we really should check that ( )2R − exists since there is always a chance that it

won’t since we are dealing with a logarithm. It does exist ( ( ) ( )2 ln 10R − = ) and so the only

critical point for this function is,

2x = −


Calculus I


17. Determine the critical points of ( ) ( )3 7ln 8 2A t t t= − + .


( ) 8 56 24 503 7 38 2 8 2 8 2

tA tt t t

− ′ = − = − = + + +

We did quite a bit of simplification of the derivative to help with the next step. While not technically required it will mean the next step will be a fair amount simpler to do. Step 2 Recall that critical points are simply where the derivative is zero and/or doesn’t exist. In this case the derivative is a rational expression. So, we know that the derivative will be zero if the numerator is zero (and the denominator is also not zero for the same values of course). We also know that the derivative won’t exist if we get division by zero. However, in this case note that the denominator is also the polyomial that is inside the logarithm and so any values of t for which the denominator is zero (i.e. 1

4t = − since it’s easy to see that point) will not be in the

domain of the original function (i.e. the function, ( )14A − , won’t exist because we can’t take the

logarithm of zero). Therefore, this point will not be a critical point. So, setting the numerator equal to zero gives, 25

1224 50 0t t− = ⇒ =

Step 3 As a final step we really should check that ( )25

12A exists since there is always a chance that it

won’t since we are dealing with a logarithm. It does exist ( ( ) ( )25 75 6512 12 37lnA = − ) and so the

only critical point for this function is,

2512t =


Calculus I




Solution There really isn’t all that much to this problem. We know that absolute extrema are the highest/lowest point on the graph and that they may occur at the endpoints or in the interior of the graph. Relative extrema on the other hand, are “humps” or “bumps” in the graph where in the region around that point the “bump” is a maximum or minimum. Also recall that relative extrema only occur in the interior of the graph and not at the end points of the interval. Also recall that relative extrema can also be absolute extrema. So, we have the following absolute/relative extrema. Absolute Maximum : ( )4,5

Absolute Minimum : ( )2, 6−

Relative Maximums : ( )1, 2− and ( )4,5

Relative Minimums : ( )3, 2− − and ( )2, 6−


Calculus I




Solution There really isn’t all that much to this problem. We know that absolute extrema are the highest/lowest point on the graph and that they may occur at the endpoints or in the interior of the graph. Relative extrema on the other hand, are “humps” or “bumps” in the graph where in the region around that point the “bump” is a maximum or minimum. Also recall that relative extrema only occur in the interior of the graph and not at the end points of the interval. Also recall that relative extrema can also be absolute extrema. So, we have the following absolute/relative extrema. Absolute Maximum : ( )6,8

Absolute Minimum : ( )9, 6−

Relative Maximums : ( )1,3 and ( )6,8

Relative Minimums : ( )2, 1− − and ( )2, 4−

3. Sketch the graph of ( ) 2 4g x x x= − and identify all the relative extrema and absolute extrema



Calculus I


(b) [ ]1, 4−

(c) [ ]1,3

(d) [ ]3,5

(e) ( ]1,5−

(a) ( ),−∞ ∞

Here’s a graph of the function on the interval.

If you don’t recall how to graph parabolas you should check out the section on graphing parabolas in the Algebra notes. So, on the interval ( ),−∞ ∞ , we can clearly see that there are no absolute maximums (the graph

increases without bounds on both the left and right side of the graph.). There are also no relative maximums (there are no “bumps” in which the graph is a maximum in the region around the point). The point ( )2, 4− is both a relative minimum and an absolute minimum.

(b) [ ]1, 4−

Here’s a graph of the function on this interval.


Calculus I


The point ( )2, 4− is still both a relative minimum and an absolute minimum. There are still

no relative maximums. However, because we are now working on a closed interval (i.e. we are working on an interval with finite endpoints and we are including the endpoints) we can see that we have an absolute maximum at the point ( )1,5− .

(c) [ ]1,3


The point ( )2, 4− is still both a relative minimum and an absolute minimum. There are still

no relative maximums of the function on this interval. However, because we are now working on a closed interval (i.e. we are working on an interval with finite endpoints and we are including the endpoints) we can see that we have an absolute maximum that occurs at the points ( )1, 3−

and ( )3, 3− .

Recall that while there can only be one absolute maximum value of a function (or minimum value if that is the case) it can occur at more than one point. (d) [ ]3,5



Calculus I


On this interval we clearly do not have any “bumps” in the interior of the interval and so, for this interval, there are no relative extrema of the function on this interval. However, we are working on a closed interval and so we can clearly see that there is an absolute maximum at the point

( )5,5 and an absolute minimum at the point ( )3, 3− .

(e) ( ]1,5−


The point ( )2, 4− is both a relative minimum and an absolute minimum. There are no

relative maximums of the function on this interval. For the absolute maximum we need to be a little careful however. In this case we are including the right endpoint of the interval, but not the left endpoint. Therefore, there is an absolute maximum at the point ( )5,5 . There is not, however, an absolute maximum at the left point

because that point is not being included in the interval. Because we are not including the left endpoint in the interval and so x will get closer and closer to

1x = − without actually reaching 1x = − . This means that while the graph will get closer and closer to 5y = it will never actually reach 5y = and so there will not be an absolute maximum at the left end point.

4. Sketch the graph of ( ) ( )34h x x= − + and identify all the relative extrema and absolute


(b) [ ]5.5, 2− −

(c) [ )4, 3− −


Calculus I


(d) [ ]4, 3− −

(a) ( ),−∞ ∞

Here’s a graph of the function on the interval.

To graph this recall the transformations of graphs. The “-” in front simply reflects the graph of

3x about the x-axis and the “+4” shifts that graph 4 units to the left. So, on the interval ( ),−∞ ∞ , we can clearly see that there are no absolute maximums (the graph

increases/decreases without bounds on both the left/right side of the graph.). There are also no relative extrema (there are no “bumps” in which the graph is a maximum or minimum in the region around the point). Don’t get so locked into functions having to have extrema of some kind. There are all sorts of graphs that do not have absolute or relative extrema. This is one of those. (b) [ ]5.5, 2− −



Calculus I


As with the first part we still have no relative extrema. However, because we are now working on a closed interval (i.e. we are working on an interval with finite endpoints and we are including the endpoints) we can see that we will have absolute extrema in the interval. We will have an absolute maximum at the point ( )5.5, 3.375− and an absolute minimum at

the point ( )2, 8− − .

(c) [ )4, 3− −


We still have no relative extrema for this function. Because we are including the left endpoint in the interval we can see that we have an absolute maximum at the point ( )4,0− .

We need to be careful with the right endpoint however. It may look like we have an absolute minimum at that point, but we don’t. We are not including 3x = − in our interval. What this means is that we are going to continue to take values of x that are closer and closer to 3x = − and graphing them, but we aren’t going to ever reach 3x = − . Therefore, technically, the graph will continually decreases without ever actually reaching a final value. It will get closer and closer to -1, but will never actually reach that point. What this means for us is that there will be no absolute minimum of the function on the given interval. (d) [ ]4, 3− −

Here is a graph of the function on this interval.


Calculus I


Note that the only difference between this part and the previous part is that we are now including the right endpoint in the interval. Because of that most of the answers here are identical to part (c). There are no relative extrema of the function on the interval and there is an absolute maximum at the point ( )4,0− .

Now, unlike part (c) we are including 3x = − in the interval and so the graph will reach a final point, so to speak, as we move to the right. Therefore, for this interval, we have an absolute minimum at the point ( )3, 1− − .

5. Sketch the graph of some function on the interval [ ]1,6 that has an absolute maximum at

6x = and an absolute minimum at 3x = . Hint :Do not let the apparent difficulty of this problem fool you. It’s not asking us to find an actual function that meets these conditions. It’s only asking for a graph that meets the conditions and we know what absolute extrema look like so just start sketching and keep in mind what the conditions are. Step 1 So, we need a graph of some function (not the function itself, only the graph). The graph must be on the interval [ ]1,6 and must have absolute extrema at the specified points.

By this point we should have seen enough sketches of graphs to have a pretty good idea of what absolute minimums that are not at the endpoints of an interval should look like on a graph. Therefore, we should know basically what the graph should look like at 3x = .


Calculus I


Next, we know that the absolute maximum must occur at the right end point of the interval and so all we need to do is sketch a curve from the absolute minimum up to the right endpoint and make sure that the graph at the right endpoint is simply higher than every other point on the graph. For the graph to the left of the absolute minimum we can sketch in pretty much anything until we reach the left end point, we just need to make sure that no portion of it goes below the absolute minimum or above the absolute maximum. Step 2 There are literally an infinite number of graphs that we could do here. Some will be more complicated that others, but here is probably one of the simpler graphs that we could use here.

6. Sketch the graph of some function on the interval [ ]4,3− that has an absolute maximum at

3x = − and an absolute minimum at 2x = . Hint :Do not let the apparent difficulty of this problem fool you. It’s not asking us to find an actual function that meets these conditions. It’s only asking for a graph that meets the conditions and we know what absolute extrema look like so just start sketching and keep in mind what the conditions are. Step 1 So, we need a graph of some function (not the function itself, only the graph). The graph must be on the interval [ ]4,3− and must have absolute extrema at the specified points.

By this point we should have seen enough sketches of graphs to have a pretty good idea of what absolute maximums/minimums that are not at the endpoints of an interval should look like on a graph. Therefore, we should know basically what the graph should look like at 3x = − and


Calculus I


2x = . There are many ways we could sketch the graph between these two points, but there is no reason to overly complicate the graph so the best thing to do is probably just sketch in a short smooth curve connecting the two points. Also, because the absolute extrema occur interior to the interval we know that the graph at the endpoints of the interval must fall somewhere between the maximum/minimum values of the graph. This means that as we sketch the graph from the absolute maximum to the left end point we can sketch anything we just need to make sure it never rises above the highest point on the graph or below the lowest point on the graph. Similarly as we sketch the graph from the absolute minimum to the right endpoint we just need to make sure it stays between the highest and lowest point on the graph. Step 2 There are literally an infinite number of graphs that we could do here. Some will be more complicated that others, but here is probably one of the simpler graphs that we could use here.

7. Sketch the graph of some function that meets the following conditions :

(a) The function is continuous. (b) Has two relative minimums. (b) One of relative minimums is also an absolute minimum and the other relative

minimum is not an absolute minimum. (c) Has one relative maximum. (d) Has no absolute maximum.

Hint :Do not let the apparent difficulty of this problem fool you. It’s not asking us to find an actual function that meets these conditions. It’s only asking for a graph that meets the conditions


Calculus I


and we know what absolute and relative extrema look like so just start sketching and keep in mind what the conditions are. Step 1 So, we need a graph of some function (not the function itself, only the graph) that meets the given conditions. We were not given an interval as one of the conditions so it’s okay to assume that the interval is ( ),−∞ ∞ for this problem.

From the first condition we know that we can’t have any holes or breaks in the graph in order for the function to be continuous. Now let’s take care of the next two conditions as they are related to each other. By this point we’ve seen enough sketches of graphs to have a pretty good idea of what absolute and relative minimums looks like. So, we’re going to need two downwards pointing “bumps” in the graph to give use the two relative minimums. Also, one of them must be the lowest point on the graph and other must be higher so it is not also an absolute minimum. Next, we want to think about how to connect the two relative minimums. This is also where the fourth condition comes in. As we’ll see because we have a continuous function we’ll need that to connect the two relative minimums. Let’s start with the leftmost relative minimum. In order for it to be a minimum the graph must be increasing as we move to the right. However, if we also want to get the minimum to the right of this the graph will have to, at some point, start decreasing again. If you think about it that is exactly what a relative maximum will look like. So, in moving from the leftmost relative minimum to the rightmost relative minimum we must have a relative maximum between them and so the fourth condition is automatically met. Note that if we don’t insist on a continuous function it is possible to get from one to the other without having a relative minimum. All it would take is to have a division by zero discontinuity somewhere between the two relative minimums in which the graph goes to positive infinity on both sides of the discontinuity. This would maintain the relative minimums and at the same time would not be a relative maximum. Now let’s deal with the final condition. In order for the graph to have no absolute maximum all we really need to do is make sure that the graph increases without bound as we move to the right and left of the graph. This will also match up nicely with the relative minimums that we are required to have. To the left of the leftmost relative minimum the graph must be increasing and so we may as well just let it increase forever on that side. Likewise, on the right side of the rightmost relative


Calculus I


minimum the graph will need to be increasing. So, again let’s just let the graph increase forever on that side. Step 2 There are literally an infinite number of graphs that we could do here. Some will be more complicated that others, but here is probably one of the simpler graphs that we could use here.

Finding Absolute Extrema 1. Determine the absolute extrema of ( ) 3 28 81 42 8f x x x x= + − − on [ ]8,2− .

Hint : Just recall the process for finding absolute extrema outlined in the notes for this section and you’ll be able to do this problem! Step 1 First, notice that we are working with a polynomial and this is continuous everywhere and so will be continuous on the given interval. Recall that this is important because we now know that absolute extrema will in fact exist by the Extreme Value Theorem! Now that we know that absolute extrema will in fact exist on the given interval we’ll need to find the critical points of the function. Given that the purpose of this section is to find absolute extrema we’ll not be putting much work/explanation into the critical point steps. If you need practice finding critical points please go back and work some problems from that section.


Calculus I


Here are the critical points for this function. ( ) ( )( )2 1

424 164 42 6 4 1 7 0 7,f x x x x x x x′ = + − = − + = ⇒ = − = Step 2 Now, recall that we actually are only interested in the critical points that are in the given interval and so, in this case, the critical points that we need are, 1

47,x x= − = Step 3 The next step is to evaluate the function at the critical points from the second step and at the end points of the given interval. Here are those function evaluations. ( ) ( ) ( ) ( )1

48 1416 7 1511 13.3125 2 296f f f f− = − = = − = Do not forget to evaluate the function at the end points! This is one of the biggest mistakes that people tend to make with this type of problem. Step 4 The final step is to identify the absolute extrema. So, the answers for this problem are then,

14

Absolute Maximum : 1511 at 7Absolute Minimum : 13.3125 at

xx

= −− =

2. Determine the absolute extrema of ( ) 3 28 81 42 8f x x x x= + − − on [ ]4,2− .

Hint : Just recall the process for finding absolute extrema outlined in the notes for this section and you’ll be able to do this problem! Step 1 First, notice that we are working with a polynomial and this is continuous everywhere and so will be continuous on the given interval. Recall that this is important because we now know that absolute extrema will in fact exist by the Extreme Value Theorem! Now that we know that absolute extrema will in fact exist on the given interval we’ll need to find the critical points of the function.


Calculus I


Given that the purpose of this section is to find absolute extrema we’ll not be putting much work/explanation into the critical point steps. If you need practice finding critical points please go back and work some problems from that section. Here are the critical points for this function. ( ) ( )( )2 1

424 164 42 6 4 1 7 0 7,f x x x x x x x′ = + − = − + = ⇒ = − = Step 2 Now, recall that we actually are only interested in the critical points that are in the given interval and so, in this case, the only critical point that we need is, 1

4x = Step 3 The next step is to evaluate the function at the critical point from the second step and at the end points of the given interval. Here are those function evaluations. ( ) ( ) ( )1

44 944 13.3125 2 296f f f− = = − = Do not forget to evaluate the function at the end points! This is one of the biggest mistakes that people tend to make with this type of problem. Step 4 The final step is to identify the absolute extrema. So, the answers for this problem are then,

14

Absolute Maximum : 944 at 4Absolute Minimum : 13.3125 at

xx

= −− =

Note the importance of paying attention to the interval with this problem. Had we neglected to exclude 7x = − we would have gotten the wrong answer for the absolute maximum (check out the previous problem to see this….). 3. Determine the absolute extrema of ( ) 3 4 51 80 5 2R t t t t= + + − on [ ]4.5, 4− .

Hint : Just recall the process for finding absolute extrema outlined in the notes for this section and you’ll be able to do this problem! Step 1


Calculus I


First, notice that we are working with a polynomial and this is continuous everywhere and so will be continuous on the given interval. Recall that this is important because we now know that absolute extrema will in fact exist by the Extreme Value Theorem! Now that we know that absolute extrema will in fact exist on the given interval we’ll need to find the critical points of the function. Given that the purpose of this section is to find absolute extrema we’ll not be putting much work/explanation into the critical point steps. If you need practice finding critical points please go back and work some problems from that section. Here are the critical points for this function. ( ) ( )( )2 3 4 2240 20 10 10 6 4 0 4, 0, 6R t t t t t t t t t t′ = + − = − − + = ⇒ = − = = Step 2 Now, recall that we actually are only interested in the critical points that are in the given interval and so, in this case, the critical points that we need are, 4, 0t t= − = Step 3 The next step is to evaluate the function at the critical points from the second step and at the end points of the given interval. Here are those function evaluations. ( ) ( ) ( ) ( )4.5 1548.13 4 1791 0 1 4 4353R R R R− = − − = − = = Do not forget to evaluate the function at the end points! This is one of the biggest mistakes that people tend to make with this type of problem. Step 4 The final step is to identify the absolute extrema. So, the answers for this problem are then,

Absolute Maximum : 4353 at 4Absolute Minimum : 1791 at 4

tt=

− = −

Note the importance of paying attention to the interval with this problem. Had we neglected to exclude 6t = we would have gotten the wrong answer for the absolute maximum. Also note that if we’d neglected to check the endpoints at all we also would have gotten the wrong absolute maximum.


Calculus I


4. Determine the absolute extrema of ( ) 3 4 51 80 5 2R t t t t= + + − on [ ]0,7 .

Hint : Just recall the process for finding absolute extrema outlined in the notes for this section and you’ll be able to do this problem! Step 1 First, notice that we are working with a polynomial and this is continuous everywhere and so will be continuous on the given interval. Recall that this is important because we now know that absolute extrema will in fact exist by the Extreme Value Theorem! Now that we know that absolute extrema will in fact exist on the given interval we’ll need to find the critical points of the function. Given that the purpose of this section is to find absolute extrema we’ll not be putting much work/explanation into the critical point steps. If you need practice finding critical points please go back and work some problems from that section. Here are the critical points for this function. ( ) ( )( )2 3 4 2240 20 10 10 6 4 0 4, 0, 6R t t t t t t t t t t′ = + − = − − + = ⇒ = − = = Step 2 Now, recall that we actually are only interested in the critical points that are in the given interval and so, in this case, the critical points that we need are, 0, 6t t= = Do not get excited about the fact that one of the critical points also happens to be one of the end points of the interval. This happens on occasion. Step 3 The next step is to evaluate the function at the critical points from the second step and at the end points of the given interval. Here are those function evaluations. ( ) ( ) ( )0 1 6 8209 7 5832R R R= = = Do not forget to evaluate the function at the end points! This is one of the biggest mistakes that people tend to make with this type of problem. Step 4 The final step is to identify the absolute extrema. So, the answers for this problem are then,


Calculus I



tt

==

Note the importance of paying attention to the interval with this problem. Had we neglected to exclude 4t = − we would have gotten the wrong answer for the absolute minimum. 5. Determine the absolute extrema of ( ) 3 24 3 9 12h z z z z= − + + on [ ]2,1− .

Hint : Just recall the process for finding absolute extrema outlined in the notes for this section and you’ll be able to do this problem! Step 1 First, notice that we are working with a polynomial and this is continuous everywhere and so will be continuous on the given interval. Recall that this is important because we now know that absolute extrema will in fact exist by the Extreme Value Theorem! Now that we know that absolute extrema will in fact exist on the given interval we’ll need to find the critical points of the function. Given that the purpose of this section is to find absolute extrema we’ll not be putting much work/explanation into the critical point steps. If you need practice finding critical points please go back and work some problems from that section. Here are the critical points for this function.

( ) 2 6 396 1 1112 6 9 024 4

ih z z z z ± − ±′ = − + = ⇒ = = Now, recall that we only work with real numbers here and so we ignore complex roots. Therefore this function has no critical points. Step 2 Technically the next step is to determine all the critical points that are in the given interval. However, there are no critical points for this function and so there are also no critical points in the given interval. Step 3 The next step is to evaluate the function at the critical points from the second step and at the end points of the given interval. However, since there are no critical points for this function all we need to do is evaluate the function at the end points of the interval.


Calculus I


Here are those function evaluations. ( ) ( )2 50 1 22h h− = − = Do not forget to evaluate the function at the end points! This is one of the biggest mistakes that people tend to make with this type of problem. That is especially true for this problem as there would be no points to evaluate at without the end points. Step 4 The final step is to identify the absolute extrema. So, the answers for this problem are then,


zz=

− = −

Note that if we hadn’t remembered to evaluate the function at the end points of the interval we would not have had an answer for this problem! 6. Determine the absolute extrema of ( ) 4 3 23 26 60 11g x x x x= − + − on [ ]1,5 .

Hint : Just recall the process for finding absolute extrema outlined in the notes for this section and you’ll be able to do this problem! Step 1 First, notice that we are working with a polynomial and this is continuous everywhere and so will be continuous on the given interval. Recall that this is important because we now know that absolute extrema will in fact exist by the Extreme Value Theorem! Now that we know that absolute extrema will in fact exist on the given interval we’ll need to find the critical points of the function. Given that the purpose of this section is to find absolute extrema we’ll not be putting much work/explanation into the critical point steps. If you need practice finding critical points please go back and work some problems from that section. Here are the critical points for this function. ( ) ( ) ( )3 2 5

212 78 120 6 4 2 5 0 0, , 4g x x x x x x x x x x′ = − + = − − = ⇒ = = = Step 2


Calculus I


Now, recall that we actually are only interested in the critical points that are in the given interval and so, in this case, the critical points that we need are, 5

2 , 4x x= = Step 3 The next step is to evaluate the function at the critical points from the second step and at the end points of the given interval. Here are those function evaluations. ( ) ( ) ( ) ( )5

21 26 74.9375 4 53 5 114g g g g= = = = Do not forget to evaluate the function at the end points! This is one of the biggest mistakes that people tend to make with this type of problem. Step 4 The final step is to identify the absolute extrema. So, the answers for this problem are then,


xx

==

Note that if we hadn’t remembered to evaluate the function at the end points of the interval we would have gotten both of the answers incorrect!

7. Determine the absolute extrema of ( ) ( ) ( )34 22 8 9Q x x x= − − on [ ]3,3− .

Hint : Just recall the process for finding absolute extrema outlined in the notes for this section and you’ll be able to do this problem! Step 1 First, notice that we are working with a polynomial and this is continuous everywhere and so will be continuous on the given interval. Recall that this is important because we now know that absolute extrema will in fact exist by the Extreme Value Theorem! Now that we know that absolute extrema will in fact exist on the given interval we’ll need to find the critical points of the function. Given that the purpose of this section is to find absolute extrema we’ll not be putting much work/explanation into the critical point steps. If you need practice finding critical points please go back and work some problems from that section.


Calculus I


Here are the critical points for this function.

( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) ( )

3 23 42 2

23 2 2

3 579614 40

4 8 2 8 9 3 2 2 8 9

4 2 8 9 20 3 72

0 , 3, 1.8239, 1.9739

Q x x x x x x

x x x x

x x x ±

′ = − − − + − −

= − − − − −

= ⇒ = = ± = = −

Step 2 Now, recall that we actually are only interested in the critical points that are in the given interval and so, in this case, we need all the critical points from the first step.

3 579614 40, 3, 1.8239, 1.9739x x x ±= = ± = = −

Do not get excited about the fact that both end points of the interval are also critical points. It happens sometimes and in this case it will reduce the number of computations required in the next step by 2 and that’s not a bad thing. Step 3 The next step is to evaluate the function at the critical points from the second step and at the end points of the given interval. Here are those function evaluations. ( ) ( ) ( ) ( ) ( )7 61

43 0 1.8239 1.38 10 0 1.9739 4.81 10 3 0Q Q Q Q Q− = − = − × = = − × = Do not get excited about the large numbers for the two non-zero function values. This is something that is going to happen on occasion and we shouldn’t worry about it when it does happen. Step 4 The final step is to identify the absolute extrema. So, the answers for this problem are then,

14

7

Absolute Maximum : 0 at 3, , 3

Absolute Minimum : 1.38 10 at 1.8239

x x x

x

= − = =

− × = − Recall that while we can only have one largest possible value (i.e. only one absolute maximum) it is completely possible for it to occur at more than one point (3 points in this case).

8. Determine the absolute extrema of ( ) ( )532 2h w w w= + on 5 12 2,− .


Calculus I


Hint : Just recall the process for finding absolute extrema outlined in the notes for this section and you’ll be able to do this problem! Step 1 First, notice that we are working with a polynomial and this is continuous everywhere and so will be continuous on the given interval. Recall that this is important because we now know that absolute extrema will in fact exist by the Extreme Value Theorem! Now that we know that absolute extrema will in fact exist on the given interval we’ll need to find the critical points of the function. Given that the purpose of this section is to find absolute extrema we’ll not be putting much work/explanation into the critical point steps. If you need practice finding critical points please go back and work some problems from that section. Here are the critical points for this function.

( ) ( ) ( )

( ) ( )

5 42 3

52 34

6 2 10 2

4 2 4 3 0 0, , 2

h w w w w w

w w w w w w

′ = + + +

= + + = ⇒ = = − = −

Step 2 Now, recall that we actually are only interested in the critical points that are in the given interval and so, in this case, we need all the critical points from the first step. 3

40, , 2w w w= = − = − Step 3 The next step is to evaluate the function at the critical points from the second step and at the end points of the given interval. Here are those function evaluations. ( ) ( ) ( ) ( ) ( )5 3 1

2 4 20.9766 2 0 2.5749 0 0 24.4141h h h h h− = − = − = − = = Step 4 The final step is to identify the absolute extrema. So, the answers for this problem are then,

12

34

Absolute Maximum : 24.4141 at Absolute Minimum : 2.5749 at

ww

=

− = −


Calculus I


9. Determine the absolute extrema of ( ) 2

42 8

zf zz z

+=

+ + on [ ]10,0− .

Hint : Just recall the process for finding absolute extrema outlined in the notes for this section and you’ll be able to do this problem! Step 1 First, notice that we are working with a rational expression in which both the numerator and denominator are continuous everywhere. Also notice that the rational expression exists at all points in the interval and so will be continuous on the given interval. Recall that this is important because we now know that absolute extrema will in fact exist by the Extreme Value Theorem! Now that we know that absolute extrema will in fact exist on the given interval we’ll need to find the critical points of the function. Given that the purpose of this section is to find absolute extrema we’ll not be putting much work/explanation into the critical point steps. If you need practice finding critical points please go back and work some problems from that section. Here are the critical points for this function.

( )( )( ) ( ) ( )

( )( )

( )

2

22

28 72

222

1 2 8 4 4 1

2 8

2 8 20 4 3 2 8.2426, 0.2426

2 8

z z z zf z

z z

z zz

z z− ±

+ + − + +′ =

+ +

− + −= = ⇒ = = − ± = −

+ +

Step 2 Now, recall that we actually are only interested in the critical points that are in the given interval and so, in this case, the only critical point that we need is,

4 3 2 8.2426z = − − = − Step 3 The next step is to evaluate the function at the critical point from the second step and at the end points of the given interval. Here are those function evaluations. ( ) ( ) ( )1 1

33 210 0.0303 8.2426 0.03128 0f f f− = − = − − = − = Step 4 The final step is to identify the absolute extrema. So, the answers for this problem are then,


Calculus I


12Absolute Maximum : at 0

Absolute Minimum : 0.03128 at 4 3 2

z

z

=

− = − − Note the importance of paying attention to the interval with this problem. Had we neglected to

exclude 4 3 2 0.2426z = − + = we would have gotten the wrong answer for the absolute maximum. This problem also shows that we need to be very careful with doing too much rounding of our answers. Had we rounded down to say 2 decimal places we would have been tempted to say that the absolute minimum occurred at two places when in fact one of the points was lower than the other.

10. Determine the absolute extrema of ( ) ( )2

2 310A t t t= − on [ ]2, 10.5 .

Hint : Just recall the process for finding absolute extrema outlined in the notes for this section and you’ll be able to do this problem! Step 1 First, notice that we are working with a product of a polynomial and a cube root function. Bother are continuous everywhere and so the product will be continuous on the given interval. Recall that this is important because we now know that absolute extrema will in fact exist by the Extreme Value Theorem! Now that we know that absolute extrema will in fact exist on the given interval we’ll need to find the critical points of the function. Given that the purpose of this section is to find absolute extrema we’ll not be putting much work/explanation into the critical point steps. If you need practice finding critical points please go back and work some problems from that section. Here are the critical points for this function.

( ) ( ) ( )( ) ( ) ( )( )

( )( ) ( )

( )( )

22 1 22 23 3 3

3 13

2 2

1 1 13 3 3

152

22 10 1 10 2 103 10

6 10 2 4 15 260 8

3 10 3 10 3 100 0, , 10

tA t t t t t t tt

t t t t tt t

t t tt t t

−′ = − + − − = − −−

− − −−= = =

− − −

= = = =


Calculus I


Don’t forget about critical points where the derivative doesn’t exist! Step 2 Now, recall that we actually are only interested in the critical points that are in the given interval and so, in this case, the critical points that we need are, 15

2 , 10t t= = Step 3 The next step is to evaluate the function at the critical points from the second step and at the end points of the given interval. Here are those function evaluations. ( ) ( ) ( ) ( )15

22 16 103.613 10 0 10.5 69.4531A A A A= = = = Step 4 The final step is to identify the absolute extrema. So, the answers for this problem are then,

152Absolute Maximum : 103.613 at

Absolute Minimum : 0 at 10t

t=

=

Note the importance of paying attention to the interval with this problem. Had we neglected to exclude 0t = we would have had the absolute minimum showing up at two places instead of only the one place inside the given interval.

11. Determine the absolute extrema of ( ) ( )3 92sin y yf y = + on [ ]10,15− .

Hint : Just recall the process for finding absolute extrema outlined in the notes for this section and you’ll be able to do this problem! Step 1 First, notice that we are working with the sine function and this is continuous everywhere and so will be continuous on the given interval. Recall that this is important because we now know that absolute extrema will in fact exist by the Extreme Value Theorem! Now that we know that absolute extrema will in fact exist on the given interval we’ll need to find the critical points of the function.


Calculus I


Given that the purpose of this section is to find absolute extrema we’ll not be putting much work/explanation into the critical point steps. If you need practice finding critical points please go back and work some problems from that section. Here are the critical points for this function.

( ) ( ) ( ) ( )11 2 23 3 9 3 3 3 3

2cos 0 cos cos 2.3005y y yf y −′ = + = → = − → = − =

3

3

2.3005 2 6.9016 60, 1, 2, ,

11.9481 63.9827 2

y

yn y n

ny nn

π πππ

= + = +⇒ = ± ± ±

= += +K

If you need some review on solving trig equations please go back to the Review chapter and work some of the problems the Solving Trig Equations sections. Step 2 Now, recall that we actually are only interested in the critical points that are in the given interval and so, in this case, the critical points that we need are, 6.9016, 6.9016, 11.9481y y y= − = = Note that we got these values by plugging in values of n into the solutions above and checking the results against the given interval. Step 3 The next step is to evaluate the function at the critical points from the second step and at the end points of the given interval. Here are those function evaluations.

( ) ( ) ( )

( ) ( )10 2.0317 6.9016 2.2790 6.9016 2.2790

11.9481 1.9098 15 2.3744

f f f

f f

− = − − = − =

= =

Step 4 The final step is to identify the absolute extrema. So, the answers for this problem are then,

Absolute Maximum : 2.3744 at 15Absolute Minimum : 2.2790 at 6.9016

yy=

− = −

Note the importance of paying attention to the interval with this problem. Without an interval we would have had (literally) an infinite number of critical points to check. Also, without an interval (as a quick graph of the function would show) there would be no absolute extrema for this function.


Calculus I


12. Determine the absolute extrema of ( ) 3 22 7w w wg w − −= e on 512 2,− .

Hint : Just recall the process for finding absolute extrema outlined in the notes for this section and you’ll be able to do this problem! Step 1 First, notice that we are working with an exponential function with a polynomial in the exponent. The exponent is continuous everywhere and so we can see that the exponential function will also be continuous everywhere. Therefore the function will be continuous on the given interval. Recall that this is important because we now know that absolute extrema will in fact exist by the Extreme Value Theorem! Now that we know that absolute extrema will in fact exist on the given interval we’ll need to find the critical points of the function. Given that the purpose of this section is to find absolute extrema we’ll not be putting much work/explanation into the critical point steps. If you need practice finding critical points please go back and work some problems from that section. Here are the critical points for this function.

( ) ( )

( )( )

3 2

3 2

2 2 7

2 7 73

3 4 7

1 3 7 0 1,

w w w

w w w

g w w w

w w w w

− −

− −

′ = − −

= + − = ⇒ = − =

e

e

Step 2 Now, recall that we actually are only interested in the critical points that are in the given interval and so, in this case, the only critical point that we need is, 7

3w = Step 3 The next step is to evaluate the function at the critical point from the second step and at the end points of the given interval. Here are those function evaluations.

( ) ( ) ( )23 392 1158 27 87 51

2 3 2g g g− −− = = =e e e Step 4 The final step is to identify the absolute extrema. So, the answers for this problem are then,


Calculus I


238

39227

12

73

Absolute Maximum : at

Absolute Minimum : at

w

w−

= −

=

e

e Note the importance of paying attention to the interval with this problem. Had we neglected to exclude 1w = − we would have gotten the absolute maximum wrong. Also note that we need to be careful with rounding with this problem. Both of the exponentials with negative exponents are very small and rounding could cause some real issues here. However, we don’t need to actually do any calculator work for this anyway. Recall that the more negative the exponent is the smaller the exponential will be.

So, because 392 11527 8> we must have

392 11527 8− −<e e .

13. Determine the absolute extrema of ( ) ( )2ln 4 14R x x x= + + on [ ]4,2− .

Hint : Just recall the process for finding absolute extrema outlined in the notes for this section and you’ll be able to do this problem! Step 1 First, notice that we are working with a logarithm whose argument is a polynomial (which is continuous everywhere) that is always positive in the interval. Because of this we can see that the function will be continuous on the given interval. Recall that this is important because we now know that absolute extrema will in fact exist by the Extreme Value Theorem! Now that we know that absolute extrema will in fact exist on the given interval we’ll need to find the critical points of the function. Given that the purpose of this section is to find absolute extrema we’ll not be putting much work/explanation into the critical point steps. If you need practice finding critical points please go back and work some problems from that section. Here are the critical points for this function.

( ) 2

2 4 24 14xR x x

x x+′ = ⇒ = −

+ +

Step 2 Now, recall that we actually are only interested in the critical points that are in the given interval and so, in this case, the critical point that we need is,


Calculus I


2x = − Step 3 The next step is to evaluate the function at the critical points from the second step and at the end points of the given interval. Here are those function evaluations. ( ) ( ) ( )4 2.6391 2 2.3026 2 3.2581R R R− = − = = Step 4 The final step is to identify the absolute extrema. So, the answers for this problem are then,

Absolute Maximum : 3.2581 at 2Absolute Minimum : 2.3026 at 2

xx

== −

The Shape of a Graph, Part I 1. The graph of a function is given below. Determine the open intervals on which the function increases and decreases.

Solution There really isn’t too much to this problem. We can easily see from the graph where the function in increasing/decreasing and so all we need to do is write down the intervals.

( ) ( ) ( ) ( )Increasing : 3,1 & 7, Decreasing : , 3 & 1,7− ∞ −∞ −


Calculus I


Note as well that by open intervals we mean that we don’t include the end points in the interval. For this problem that is important because at the end points we are at infinity or the function is either not increasing or decreasing. 2. The graph of a function is given below. Determine the open intervals on which the function increases and decreases.

Solution There really isn’t too much to this problem. We can easily see from the graph where the function in increasing/decreasing and so all we need to do is write down the intervals.

( ) ( ) ( ) ( )Increasing : ,1 , 4,8 & 8, Decreasing : 1, 4−∞ ∞ Note as well that by open intervals we mean that we don’t include the end points in the interval. For this problem that is important because at the end points we are at infinity or the function is either not increasing or decreasing. 3. Below is the graph of the derivative of a function. From this graph determine the open intervals in which the function increases and decreases.


Calculus I


Hint : Be careful with this problem. The graph is of the derivative of the function and so we don’t just write down intervals where the graph is increasing and decreasing. Recall how the derivative tells us where the function is increasing and decreasing and this problem is not too bad. Solution We have to be careful and not do this problem as we did the first two practice problems. The graph given is the graph of the derivative and not the graph of the function. So, the answer is not just where the graph is increasing or decreasing. Instead we need to recall that the sign of the derivative tells us where the function is increasing and decreasing. If the derivative is positive (i.e. its graph is above the x-axis) then the function is increasing and if the derivative is negative (i.e. its graph is below the x-axis) then the function is decreasing. So, it is fairly clear where the graph is above/below the x-axis and so we have the following intervals of increase/decrease.

( ) ( ) ( ) ( )Increasing : 7, 2 & 2,5 Decreasing : , 7 & 5,− − − −∞ − ∞

4. This problem is about some function. All we know about the function is that it exists everywhere and we also know the information given below about the derivative of the function. Answer each of the following questions about this function.



Calculus I


( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

5 0 2 0 4 0 8 0

0 on 5, 2 , 2,4 , 8, 0 on , 5 , 4,8

f f f f

f x f x

′ ′ ′ ′− = − = = =

′ ′< − − − ∞ > −∞ −

Hint : This problem is actually quite simple. Just keep in mind how critical points are defined and how we can answer the last two parts from the derivative of the function. (a) Identify the critical points of the function. Okay, let’s recall the definition of a critical point. A critical point is any point in which the function exists and the derivative is either zero or doesn’t exist. We are given that the function exists everywhere (and in fact this part is why that in there at all….) and so we don’t really need to worry about that part of the definition for this problem. Also, from the given information about the derivative we can see that at every point the derivative is either zero, positive or negative. In other words, the derivative will exist at every point. So, all this means that the critical points of the function are those points were the derivative is zero and we are given those in the information. Therefore, the critical points of the function are,

5, 2, 4, 8x x x x= − = − = =

(b) Determine the open intervals on which the function increases and decreases. There is really not a lot to this part. We know that the function will increase where the derivative is positive and it will decrease where the derivative is negative. This positive and negative information is clearly listed above in the given information so here are the increasing/decreasing intervals for this function.

( ) ( ) ( ) ( ) ( )Increasing : , 5 & 4,8 Decreasing : 5, 2 , 2,4 & 8,−∞ − − − − ∞ (c) Classify the critical points as relative maximums, relative minimums or neither. Okay, there isn’t a lot that we need to do here. We know that relative maximums are increasing on the left and decreasing on the right and relative minimums are decreasing on the left and increasing on the right. We have all the increasing/decreasing information from the second part so here is the answer to this part.


Calculus I


5 : Relative Maximum2 : Neither

4 : Relative Minimum8 : Relative Maximum

xxxx

= −= −==

5. For ( ) 3 22 9 60f x x x x= − − answer each of the following questions.


(a) Identify the critical points of the function. We need the 1st derivative to get the critical points so here it is.

( ) ( ) ( )( )2 26 18 60 6 3 10 6 5 2f x x x x x x x′ = − − = − − = − +

Now, recall that critical points are where the derivative doesn’t exist or is zero. Clearly this derivative exists everywhere (it’s a polynomial….) and because we factored the derivative we can easily identify where the derivative is zero. The critical points of the function are,

2, 5x x= − =

(b) Determine the open intervals on which the function increases and decreases. To determine the increase/decrease information for the function all we need is a quick number line for the derivative. Here is the number line.

From this we get the following increasing/decreasing information for the function.

( ) ( ) ( )Increasing : , 2 & 5, Decreasing : 2,5−∞ − ∞ −


Calculus I


(c) Classify the critical points as relative maximums, relative minimums or neither. With the increasing/decreasing information from the previous step we can easily classify the critical points using the 1st derivative test. Here is classification of the functions critical points.

2 : Relative Maximum

5 : Relative Minimumxx

= −=

6. For ( ) 3 4 550 40 5 4h t t t t= + − − answer each of the following questions.



( ) ( ) ( )( )2 3 4 2 2 2120 20 20 20 6 20 3 2h t t t t t t t t t t′ = − − = − + − = − + −

Now, recall that critical points are where the derivative doesn’t exist or is zero. Clearly this derivative exists everywhere (it’s a polynomial….) and because we factored the derivative we can easily identify where the derivative is zero. The critical points of the function are,

3, 0, 2t t t= − = =



Calculus I



( ) ( ) ( ) ( )Increasing : 3,0 & 0, 2 Decreasing : , 3 & 2,− −∞ − ∞ (c) Classify the critical points as relative maximums, relative minimums or neither. With the increasing/decreasing information from the previous step we can easily classify the critical points using the 1st derivative test. Here is classification of the functions critical points.

3 : Relative Minimum

0 : Neither5 : Relative Maximum

ttt

= −==

7. For 3 22 10 12 12y x x x= − + − answer each of the following questions.



26 20 12dy x xdx

= − +

Now, recall that critical points are where the derivative doesn’t exist or is zero. Clearly this derivative exists everywhere (it’s a polynomial….) and because the derivative can’t be factored in this case we’ll need to do a quick quadratic formula to find where the derivative is zero. The critical points of the function are,

5 7 0.78475, 2.548583

x ±= =



Calculus I



( ) ( ) ( )5 7 5 7 5 7 5 73 3 3 3Increasing : , & , Decreasing : ,− + − +−∞ ∞

(c) Classify the critical points as relative maximums, relative minimums or neither. With the increasing/decreasing information from the previous step we can easily classify the critical points using the 1st derivative test. Here is classification of the functions critical points.

5 7

3

5 73

0.78475 : Relative Maximum

2.54858 : Relative Minimum

x

x

−

+

= =

= =

8. For ( ) ( )cos 3 2p x x x= + answer each of the following questions on 3

2 , 2− .



( ) ( )3sin 3 2p x x′ = − +

Now, recall that critical points are where the derivative doesn’t exist or is zero. Clearly this derivative exists everywhere (the sine function exists everywhere….) and so all we need to do is set the derivative equal to zero and solve. We’re not going to show all of those details so if you need to do some review of the process go back to the Solving Trig Equations sections for some examples. Here are all the critical points.


Calculus I


23

23

0.24320, 1, 2, 3,

0.8040x n

nx n

π

π

= += ± ± ±

= +K

Plugging in some n’s gives the following critical points in the interval 3

2 , 2− .

1.2904, 0.2432, 0.8040x x x= − = =



( ) ( ]

) ( )32

Increasing : 1.2904,0.2432 & 0.8040, 2

Decreasing : , 1.2904 & 0.2432,0.8040

−

− − Be careful with the end points of these intervals! We are working on the interval 3

2 , 2− and

we’ve done no work for increasing and decreasing outside of this interval and so we can’t say anything about what happens outside of the interval. (c) Classify the critical points as relative maximums, relative minimums or neither. With the increasing/decreasing information from the previous step we can easily classify the critical points using the 1st derivative test. Here is classification of the functions critical points.


0.2432 : Relative Maximum0.8040 : Relative Minimum

xxx

= −==


Calculus I


As with the last step, we need to again recall that we are only working on the interval 32 , 2− and

the classifications given here are only for those critical points in the interval. There are, of course, an infinite number of critical points outside of this interval and they can all be classified as relative minimums or relative maximums provided we do the work to justify the classifications. 9. For ( ) ( )22 5 14sin zR z z= − − answer each of the following questions on [ ]10,7− .



( ) ( )25 7 cos zR z′ = − −

Now, recall that critical points are where the derivative doesn’t exist or is zero. Clearly this derivative exists everywhere (the cosine function exists everywhere….) and so all we need to do is set the derivative equal to zero and solve. We’re not going to show all of those details so if you need to do some review of the process go back to the Solving Trig Equations sections for some examples. Here are all the critical points.

4.7328 4

0, 1, 2, 3,7.8336 4

z nn

z nππ

= += ± ± ±

= +K

Plugging in some n’s gives the following critical points in the interval [ ]10,7− .

7.8336, 4.7328, 4.7328z z z= − = − =



Calculus I



( ) ( ][ ) ( )

Increasing : 7.8336, 4.7328 & 4.7328,7

Decreasing : 10, 7.8336 & 4.7328, 4.7328

− −

− − − Be careful with the end points of these intervals! We are working on the interval [ ]10,7− and

we’ve done no work for increasing and decreasing outside of this interval and so we can’t say anything about what happens outside of the interval. (c) Classify the critical points as relative maximums, relative minimums or neither. With the increasing/decreasing information from the previous step we can easily classify the critical points using the 1st derivative test. Here is classification of the functions critical points.

7.8336 : Relative Minimum4.7328 : Relative Maximum


zzz

= −= −=

As with the last step, we need to again recall that we are only working on the interval [ ]10,7−

and the classifications given here are only for those critical points in the interval. There are, of course, an infinite number of critical points outside of this interval and they can all be classified as relative minimums or relative maximums provided we do the work to justify the classifications.

10. For ( ) 2 3 7h t t t= − answer each of the following questions.


(a) Identify the critical points of the function.


Calculus I


We need the 1st derivative to get the critical points so here it is.

( ) ( ) ( )( ) ( )( )

( )( ) ( )

( )( )

1 2 13 3 3

23

2 2 23 3 3

22 1

3

2 2

2 7 7 2 73 7

6 7 7 67 42

3 7 3 7 3 7

th t t t t t t tt

t t t t tt tt t t

−′ = − + − = − +−

− + −−= = =

− − −

Now, recall that critical points are where the derivative doesn’t exist or is zero. Because we simplified and factored the derivative as much as possible we can clearly see that the derivative does not exist at 7t = (and the function exists here…) and that the derivative is zero at 0t = and

6t = . The critical points of this function are then,

0, 6, 7t t t= = =



( ) ( ) ( ) ( )Increasing : ,0 , 6,7 & 7, Decreasing : 0,6−∞ ∞ (c) Classify the critical points as relative maximums, relative minimums or neither. With the increasing/decreasing information from the previous step we can easily classify the critical points using the 1st derivative test. Here is classification of the functions critical points.


Calculus I


0 : Relative Maximum6 : Relative Minimum7 : Neither

ttt

===

11. For ( )21

22 wf w w −= e answer each of the following questions.



( ) ( ) ( ) ( )2 2 2 21 1 1 1

2 2 2 22 2 2 22 21 1 1w w w wf w w w w w− − − −′ = − = − = − +e e e e

Now, recall that critical points are where the derivative doesn’t exist or is zero. Because we simplified and factored the derivative as much as possible we can clearly see that the derivative will exist everywhere (it’s the product of functions that exist everywhere). We can also easily see where the derivative is zero. The critical points of this function are then,

1, 1w w= − =




Calculus I


( ) ( ) ( )Increasing : 1,1 Decreasing : , 1 & 1,− −∞ − ∞ (c) Classify the critical points as relative maximums, relative minimums or neither. With the increasing/decreasing information from the previous step we can easily classify the critical points using the 1st derivative test. Here is classification of the functions critical points.

1 : Relative Minimum

1 : Relative Maximumww

= −=

12. For ( ) ( )22ln 1g x x x= − + answer each of the following questions.



( )2

2 2

2 1 41 21 1

x x xg xx x

− += − =

+ +

Now, recall that critical points are where the derivative doesn’t exist or is zero. Because we simplified and factored the derivative as much as possible we can clearly see that the derivative will exist everywhere (or at least the denominator will not be zero for any real numbers…). We’ll also need the quadratic formula to determine where the numerator, and hence the derivative, is zero. The critical points of this function are then,

2 3 0.2679, 3.7321x = ± =



Calculus I



( ) ( ) ( )Increasing : ,0.2679 & 3.7321, Decreasing : 0.2679,3.7321−∞ ∞ (c) Classify the critical points as relative maximums, relative minimums or neither. With the increasing/decreasing information from the previous step we can easily classify the critical points using the 1st derivative test. Here is classification of the functions critical points.

0.2679 : Relative Maximum3.7321 : Relative Minimum

xx

==

13. For some function, ( )f x , it is known that there is a relative maximum at 4x = . Answer

each of the following questions about this function. (a) What is the simplest form for the derivative of this function? Note : There really are many


(b) Using your answer from (a) determine the most general form of the function. (c) Given that ( )4 1f = find a function that will have a relative maximum at 4x = . Note :

You should be able to use your answer from (b) to determine an answer to this part. Hint : As noted in the problem there are many possible forms that the derivative can take. However, if we want things to remain simple just keep in mind what it takes for a point to be a critical point (why a critical point?). With that in mind it should be pretty simple to figure out a really simple form for the derivative to take to make sure we get a relative maximum at the point. (a) What is the simplest form for the derivative of this function? Note : There really are many


Okay, let’s get started with this problem.


Calculus I


The first thing that we’ll do is assume that the derivative exists everywhere. Making assumptions in a math class is generally a bad thing. However, in this case, because we are being asked to come up the form of the derivative all we are really doing here is starting that process. If we can’t find a derivative that will have a relative maximum at the point that also exists everywhere we can come back and change things up. If we can find a derivative that will exist everywhere (which we can as we’ll see) this assumption will help with keeping the derivative as simple as possible. Now, given that we are assuming that the derivative exists everywhere and we know that if we have a relative maximum at 4x = then 4x = must also be a critical point (recall Fermat’s Theorem from a couple of sections ago…). This is another reason for the assumption we made above. Fermat’s theorem requires that the derivative exist at the point in order to know that it is also a critical point. Next, because we assumed that the derivative exists everywhere (and in particular it exists at

4x = ) we know that in order for it to be a critical point we must also have ( )4 0f ′ = . There are

lots and lots of functions that will be zero at 4x = but probably one of the simplest is use,

( ) 4f x x′ = − This does give ( )4 0f ′ = as we need, however we have a problem. We can clearly see that if

4x < we would have ( ) 0f x′ < and if 4x > we would have ( ) 0f x′ > . This says that 4x =

would have to be a relative minimum and not the relative maximum that we wanted it to be. Luckily enough for us this is easy to fix. The only problem with our original guess is that the signs of the derivative to the right and left of 4x = are opposite what we need them to be. Therefore, all we need to do is change them and that can easily be done by multiplying by a negative or,

( ) 4f x x′ = −

With this choice we still have ( )4 0f ′ = and now the derivative is positive if 4x < and negative

if 4x > which means that 4x = will be a relative maximum. As noted in the problem statement there are many possible answers to this part. We will be working with the one given above. However, just to make the point here are a sampling of other derivatives all of which come from functions that have a relative maximum at 4x = . ( ) ( ) ( ) ( ) ( )2 2 416 24 18 6 1 sin 2xf x x f x x x f x f x xπ π−′ ′ ′ ′= − = + − = − = −e


Calculus I


After working the rest of this problem with ( ) 4f x x′ = − you might want to come back and see

if you can repeat the problem with one or more of these to see what you get. Hint : This is where the problem gets a little tricky (as if the previous part wasn’t tricky huh?). What we need to do here is “undo” the derivative. However, if you kept the answer from the previous part simple and you understand how differentiation works then it shouldn’t be too hard to “undo” the differentiation and determine a function that gives your derivative. Keep in mind however that the problem statement asks for the most general possible function and that is where most will run into problems with this part. (b) Using your answer from (a) determine the most general form of the function. Okay, from the previous part we have assumed that the derivative of our function is

( ) 4f x x′ = − and we need to “undo” the differentiation to determine the most general possible

function that we could have had. So, before doing this let’s recall what we know about differentiation. First, let’s recall the following formula,

( ) 1n nd x nxdx

−=

When we differentiate a power of x the power goes down by one. So, what would we have to differentiate to get x? We’ll, since the exponent on the x is 1 we would have had to differentiate an 2x . However, because we know that that the derivative of 2x is 2x and we want just an x that would mean that in fact we would have had to differentiate 21

2 x to get x. Next, also recall,

( )d kx kdx

=

So, using this as a guide is should be pretty simple to see that we would need to differentiate 4x to get 4. So, if we put these two parts together it looks like we could use the following function. ( ) 21

24f x x x= − The derivative of this function is clearly ( ) 4f x x′ = − . However, it is not the most general

possible function that gives this derivative.


Calculus I


Do not forget that the derivative of a constant is zero and so we could add any constant onto our function and get the same derivative. This is one of the biggest mistakes that students make with learning to “undo” differentiation. Any time we undo differentiation there is always the possibility that there was a constant on the original function and so we need to acknowledge that. We usually do this by adding a “ + c ” onto the end of our function. We use a general c because we have no way of knowing that the constant would be and this allows for all possible constants. Therefore, the most general function that we could use to get ( ) 4f x x′ = − is,

( ) 2124f x x x c= − +

Hint : This is the “easy” part of this problem. All we are really being asked to do is determine the specific value of c that we would need in order have the function have the value of 1 at 4x = . If you’ve reached this point of a Calculus course you should have the required Algebra knowledge (and yes it really is just Algebra) to do this part. (c) Given that ( )4 1f = find a function that will have a relative maximum at 4x = . Note : You

should be able to use your answer from (b) to determine an answer to this part. To do this part all we really need to do is plug 4x = into our answer from the previous part and set the result equal to 1. This will result in an equation with a single unknown value, c. So all we need to do then is solve for c. Here is the work for this part.

( ) ( ) ( )2121 4 4 4 4 8 1 8 7f c c c c= = − + = + → = + → = −

So, it looks like one possible function that will have a relative maximum at 4x = is,

( ) 2124 7f x x x= − −

As a final part to this problem, here is a quick graph of this function to verify that it does in fact have a relative maximum at 4x = .


Calculus I


14. Given that ( )f x and ( )g x are increasing functions. If we define ( ) ( ) ( )h x f x g x= +

show that ( )h x is an increasing function.

Hint : At first glance this problem may seem quite difficult. However, just keep in mind how we have been determining whether functions where increasing to this point and that should suggest a first step. Step 1 To this point we’ve always used the derivative to determine if a function was increasing so let’s do that here as well. Note that this may not seem all that useful because we don’t actually know what any of the functions are. However, just because we don’t know what the functions are doesn’t mean that we can’t at least write down a formula for ( )h x′ . Here is that formula.

( ) ( ) ( )h x f x g x′ ′ ′= + Hint : We were told that ( )f x and ( )g x are increasing functions so what does that tell us about

their derivatives? Step 2 We are told that both ( )f x and ( )g x are increasing functions so this means that we know that

both of their derivatives must be positive. Or, ( ) ( )0 0f x g x′ ′> >


Calculus I


Hint : Given the formula for ( )h x′ we found in Step 1 and what we noticed about the signs of

( )f x′ and ( )g x′ that we noted in Step 2 what can we say about the sign of ( )h x ?

Step 3 Okay, we are pretty much done at this point. We know from Step 1 that, ( ) ( ) ( )h x f x g x′ ′ ′= + Also from Step 2 we know that both ( )f x′ and ( )g x′ are positive. So, ( )h x′ is the sum of

two positive functions and in turn means that we must have, ( ) 0h x′ > Therefore we can see that ( )h x must be an increasing function.

-------------------------------------------------------

Final Thoughts / Strategy Discussion Figuring out how to do these kinds of problems can definitely seem quite daunting at times. That is especially true when the statement we are being asked to prove seems to be fairly "obvious" as is the case here. The sum of two increasing functions intuitively should also be increasing. The problem is that we are being asked to actually prove that and not just say "well it makes sense so it should be true". What we want to discuss here is not the proof of this fact (that is given above after all...). Instead let's take a look at the thought process that went into constructing the proof above. The first step is to really look at what we are being asked to prove. This means not just reading the statement, but reading the statement and trying to relate what we are being asked to prove to something we already know. In this case, we’re being asked to show that a general function is increasing given a set of assumptions. By this point we know how to prove that specific functions are increasing. So, let’s start with that. We know that in order to use Calculus to prove that a function is increasing we need to look at the derivative of the function. We also know that, at least symbolically, we write down the formula for the derivative of the function we are interested in for this problem. Now for the next step in the thought process. We've got a formula and we know that we need to show that it is positive. At this point we need to think about the assumptions that we were given.


Calculus I


Don't forget the assumptions. They were given for a reason and we'll need to use them. What do the assumptions tell us? How can we relate them to what we are being asked to prove? In this case, we know from the assumptions that the two derivatives were positive. For this problem this wasn't a particularly difficult step, but for other problems this can be a little tricky. Finally, we need to put the two previous thought process steps together. This can also be a fairly tricky step. If you haven't had a lot of exposure to "mathematical logic/proofs" it can be daunting to put all the information together. Often times you will need to try various ways of putting the information together before something "clicks" and you can see how to proceed. You may even need to go back to the previous step and see if there is something about the assumptions that you may have missed. In this case we could see that the derivative was the sum of two other derivatives and from our assumptions we knew that the two individual derivatives had to be positive. We also know from basic Algebra knowledge that the sum of two positive quantities has to be positive and so we are done. The key part of this whole process is that you will have to persevere. Try not to get discouraged and if something doesn't work out move on and try something else. Also, do not get so wrapped up in the process that you don't take breaks occasionally. If you keep running into road blocks then step away for a while and come back at a later point. Sometimes that is all it takes to get a fresh idea. Another thing that students often initially have difficulty with is trying to mathematically write this kind of thing down. In your mind you may have been able to see all the "logic" involved in the proof, but just couldn't see who to put it all together and write it down. If you are having that problem the best thing to do is just start writing things down. For instance, you know you need the derivative of the given function so write that down. If you don't have a specific function to differentiate can you at least symbolically write down the derivative as we did here? Once you have that written down look at the pieces and start writing down what you know about them. Actually write down what you know (i.e. things like ( ) 0f x′ > ). This seems silly at

times, but it really can help with the process. Once you have everything written down you might be able to see how to string everything together with words/explanations to prove what you want to prove.


Calculus I


15. Given that ( )f x is an increasing function and define ( ) ( ) 2h x f x= . Will ( )h x be an

increasing function? If yes, prove that ( )h x is an increasing function. If not, can you determine

any other conditions needed on the function ( )f x that will guarantee that ( )h x will also

increase? Hint : If you have trouble with these kinds of “proof” problems you might want to check out the discussion at the end of the previous problem for a “strategy” that might be useful here. This isn’t quite the same kind of problem but the strategy given there should help here as well. Hint : How do we use Calculus to determine if a function is increasing? Step 1 We know that the derivative can be used to tell us if a function is increasing so let’s find the derivative of ( )h x . Do not get excited about the fact that we don’t know what ( )f x is. We

can symbolically take the derivative with a quick application of the chain rule. Here is the derivative of ( )h x .

( ) ( ) ( ) ( ) ( )2 2h x f x f x f x f x1

′ ′ ′= = Hint : How do we use the derivative to determine if a function is increasing? Step 2 We know that a function will be increasing if its derivative is positive. So, the question we need to answer is can we guarantee that ( ) 0h x′ > if we only take into account the assumption that

( )f x is an increasing function.

From our assumption that ( )f x is an increasing function and so we know that ( ) 0f x′ > .

Now, let’s see what ( )h x′ tells us. We can see that ( )h x′ is a product of a number and two

functions. The “2” is positive and so the sign of the derivative will come from the sign of the product of ( )f x and ( )f x′ .

Hint : So, will ( )h x′ be positive?

Step 3


Calculus I


Okay, from our assumption we know that ( )f x′ is positive. However, the product isn’t

guaranteed to be positive. For example, consider ( )f x x= . Clearly, ( ) 1 0f x′ = > , and so this is an increasing function.

However, ( ) ( ) ( )( )1f x f x x x′ = = . Therefore, we can see that the product will not always be

positive. This shouldn’t be too surprising given that,

( ) ( ) [ ]2 2 2h x f x x x= = =

In this case we can clearly see that ( )h x will not always be an increasing function.

On the other had if we let ( ) xf x = e we can see that ( ) 0xf x′ = >e and we can also see that

( ) ( ) ( )( ) 2 0x x xf x f x′ = = >e e e .

So, from these two examples we can see that we can find increasing functions, ( )f x , for which

( ) 2f x may or may not always be increasing.

Hint : Can you use the examples above to determine a condition on ( )f x that will guarantee that

( )h x will be increasing?

Step 4 So, just what was the difference between the two examples above? The problem with the first example, ( )f x x= was that it wasn’t always positive and so the

product of ( )f x and ( )f x′ would not always be positive as we needed it to be.

This was not a problem with the second example however, ( ) xf x = e . In this case the function

is always positive and so the product of the function and its derivative will also be positive. That is also the added condition that we need to guarantee that ( )h x will be positive.

If we start with the assumption that ( )f x in an increasing function we need to further assume

that ( )f x is a positive function in order to guarantee that ( ) ( ) 2h x f x= will be a positive

function.


Calculus I


The Shape of a Graph, Part II 1. The graph of a function is given below. Determine the open intervals on which the function is concave up and concave down.

Solution There really isn’t too much to this problem. We can easily see from the graph where the function in concave up/concave down and so all we need to do is estimate where the concavity changes (and this really will be an estimate as it won’t always be clear) and write down the intervals.

( ) ( ) ( ) ( )Concave Up : 1,2 & 6, Concave Down : , 1 & 2,6− ∞ −∞ − Again, the endpoints of these intervals are, at best, estimates as it won’t always be clear just where the concavity changes. 2. Below is the graph the 2nd derivative of a function. From this graph determine the open intervals in which the function is concave up and concave down.


Calculus I


Hint : Be careful with this problem. The graph is of the 2nd derivative of the function and so we don’t just write down intervals where the graph is concave up and concave down. Recall how the 2nd derivative tells us where the function is concave up and concave down and this problem is not too bad. Solution We need to be careful and not do this problem as we did the first practice problem. The graph given is the graph of the 2nd derivative and not the graph of the function. Therefore, the answer is not just where the graph is concave up or concave down. What we need to do here is to recall that if the 2nd derivative is positive (i.e. the graph is above the x-axis) then the function in concave up and if the 2nd derivative is negative (i.e. the graph is below the x-axis) then the function is concave down. So, it is fairly clear where the graph is above/below the x-axis and so we have the following intervals of concave up/concave down.

( ) ( ) ( ) ( )Concave Up : , 4 , 2,3 & 3, Concave Down : 4, 2−∞ − − ∞ − − Even though the problem didn’t ask for it we can also identify that 4x = − and 2x = − are inflection points because at these points the concavity changes. Note that 3x = is not an inflection point. Clearly the 2nd derivative is zero here, but the concavity doesn’t change at this point and so it is not an inflection point. Keep in mind inflection points are only where the concavity changes and not simply where the 2nd derivative is zero. 3. For ( ) 2 312 6f x x x= + − answer each of the following questions.


Calculus I



(a) Determine a list of possible inflection points for the function. To get the list of possible inflection points for the function we’ll need the 2nd derivative of the function so here that is.

( ) ( )212 3 12 6f x x x f x x′ ′′= − = −

Now, recall that possible inflection points are where the 2nd derivative either doesn’t exist or is zero. Clearly the 2nd derivative exists everywhere (it’s a polynomial….) and, in this case, it should be fairly clear where the 2nd derivative is zero. The only possible inflection critical point of the function in this case is, 2x =

(b) Determine the open intervals on which the function is concave up and concave down. There isn’t much to this part. All we really need here is a number line for the 2nd derivative. Here that is,

From this we get the following concave up/concave down information for the function.

( ) ( )Concave Up : , 2 Concave Down : 2,−∞ ∞ (c) Determine the inflection points of the function. For this part all we need to do is interpret the results from the previous step. Recall that inflection points are points where the concavity changes (as opposed to simply the points where the 2nd derivative is zero or doesn’t exist). Therefore the single inflection point for this function is,


Calculus I


2x = 4. For ( ) 4 312 84 4g z z z z= − + + answer each of the following questions.



( ) ( ) ( )3 2 24 36 84 12 72 12 6g z z z g z z z z z′ ′′= − + = − = −

Now, recall that possible inflection points are where the 2nd derivative either doesn’t exist or is zero. Clearly the 2nd derivative exists everywhere (it’s a polynomial….) and, because we factored the 2nd derivative, it should be fairly clear where the 2nd derivative is zero. The possible inflection critical points of this function are, 0 & 6z z= =



( ) ( ) ( )Concave Up : ,0 & 6, Concave Down : 0,6−∞ ∞


Calculus I


(c) Determine the inflection points of the function. For this part all we need to do is interpret the results from the previous step. Recall that inflection points are points where the concavity changes (as opposed to simply the points where the 2nd derivative is zero or doesn’t exist). Therefore the inflection points for this function are,

0 & 6z z= = 5. For ( ) 4 3 212 6 36 2h t t t t t= + + − + answer each of the following questions.



( ) ( ) ( )3 2 2 24 36 12 36 12 72 12 12 6 1h t t t t h t t t t t′ ′′= + + − = + + = + +

Now, recall that possible inflection points are where the 2nd derivative either doesn’t exist or is zero. Clearly the 2nd derivative exists everywhere (it’s a polynomial….). In this case the 2nd derivative doesn’t factor and so we’ll need to use the quadratic formula to determine where the 2nd derivative is zero. The possible inflection critical points of this function are,

3 2 2 5.8284, 0.1716t = − ± = − −



Calculus I



( ) ( )( )

Concave Up : , 3 2 2 & 3 2 2,

Concave Down : 3 2 2, 3 2 2

−∞ − − − + ∞

− − − +

(c) Determine the inflection points of the function. For this part all we need to do is interpret the results from the previous step. Recall that inflection points are points where the concavity changes (as opposed to simply the points where the 2nd derivative is zero or doesn’t exist). Therefore the inflection points for this function are,

3 2 2 & 3 2 2t t= − − = − + 6. For ( ) ( )28 5 2 cos 3h w w w w= − + − on [ ]1, 2− answer each of the following questions.



( ) ( ) ( ) ( )5 4 3sin 3 4 9cos 3h w w w h w w′ ′′= − + + = +

Now, recall that possible inflection points are where the 2nd derivative either doesn’t exist or is zero. Clearly the 2nd derivative exists everywhere (the cosine function exists everywhere…) and so all we need to do is set the 2nd derivative equal to zero and solve. We’re not going to show all


Calculus I


of those details so if you need to do some review of the process go back to the Solving Trig Equations sections for some examples. The possible inflection critical points of this function are,

23

23

0.67710, 1, 2, 3,

1.4173w n

nw n

π

π

= += ± ± ±

= +K

Plugging in some n’s gives the following possible inflection points in the interval [ ]1, 2− .

0.6771 0.6771 1.4173w w w= − = = (b) Determine the open intervals on which the function is concave up and concave down. There isn’t much to this part. All we really need here is a number line for the 2nd derivative. Here that is,


( ) ( ][ ) ( )

Concave Up : 0.6771, 0.6771 & 1.4173, 2

Concave Down : 1, 0.6771 & 0.6771, 1.4173

−

− −

Be careful with the end points of these intervals! We are working on the interval [ ]1, 2− and

we’ve done no work for concavity outside of this interval and so we can’t say anything about what happens outside of the interval. (c) Determine the inflection points of the function. For this part all we need to do is interpret the results from the previous step. Recall that inflection points are points where the concavity changes (as opposed to simply the points where the 2nd derivative is zero or doesn’t exist). Therefore the inflection points for this function are,


Calculus I


0.6771 0.6771 1.4173w w w= − = =

As with the previous step we have to be careful and recall that we are working on the interval

[ ]1, 2− . There are infinitely many more possible inflection points and we’ve done no work

outside of the interval to determine if they are in fact inflection points!

7. For ( ) ( )234R z z z= + answer each of the following questions.



( ) ( ) ( ) ( )( )

( )( )( ) ( )( )

( )

( ) ( ) ( )( ) ( )

2 13 3

13

1 2 23 3 3

2 41 3 33

23

2

5 124 43 4

5 3 4 5 12 4 15 4 5 12 4 10 489 4 9 43 4

zR z z z zz

z z z z z z zR zz zz

− −

− +′ = + + + =+

+ − + + + − + + + ′′ = = = + ++

Note that we simplified the derivatives at each step to help with the next step. You don’t technically need to do this, but having the 2nd derivative in its “simplest” form will definitely help with getting the answer to this part. Now, recall that possible inflection points are where the 2nd derivative either doesn’t exist or is zero. Because we simplified the 2nd derivative as much as possible it is clear that the 2nd derivative won’t exist at 4z = − (and the function exists at this point as well!). It should also be clear that the 2nd derivative is zero at 48 24

10 5z = − = − . The possible inflection critical points of this function are then, 48

5 4.8 & 4z z= − = − = −

(b) Determine the open intervals on which the function is concave up and concave down.


Calculus I


There isn’t much to this part. All we really need here is a number line for the 2nd derivative. Here that is,


( ) ( ) ( )24 245 5Concave Up : , 4 & 4, Concave Down : ,− − − ∞ −∞ −

(c) Determine the inflection points of the function. For this part all we need to do is interpret the results from the previous step. Recall that inflection points are points where the concavity changes (as opposed to simply the points where the 2nd derivative is zero or doesn’t exist). Therefore the only inflection point for this function is,

245z = −

8. For ( ) 24 xh x −= e answer each of the following questions.



( ) ( ) ( )2 2 2 24 4 2 4 4 22 2 4 2 2 1x x x xh x x h x x x− − − −′ ′′= − = − + = −e e e e

Don’t forget to product rule for the 2nd derivative and factoring the exponential out will help a little with the next step.


Calculus I


Now, recall that possible inflection points are where the 2nd derivative either doesn’t exist or is zero. It should be fairly clear that the 2nd derivative exists everywhere (it is a product of two functions that exist everywhere…). We also know that exponentials are never zero and so the 2nd derivative will be zero at the solutions to 22 1 0x − = The possible inflection critical points of this function are then, 1

20.7071x = ± = ±



( ) ( ) ( )1 1 1 12 2 2 2

Concave Up : , & , Concave Down : ,−∞ − ∞ −

(c) Determine the inflection points of the function. For this part all we need to do is interpret the results from the previous step. Recall that inflection points are points where the concavity changes (as opposed to simply the points where the 2nd derivative is zero or doesn’t exist). Therefore the only inflection point for this function is,

1 12 2

0.7071 0.7071x x= − = − = =

9. For ( ) 5 45 8g t t t= − + answer each of the following questions.

(a) Identify the critical points of the function. (b) Determine the open intervals on which the function increases and decreases.


Calculus I


(c) Classify the critical points as relative maximums, relative minimums or neither. (d) Determine the open intervals on which the function is concave up and concave down. (e) Determine the inflection points of the function. (f) Use the information from steps (a) – (e) to sketch the graph of the function.

(a) Identify the critical points of the function. The parts to this problem (with the exception of the last part) are just like the basic increasing/decreasing problems from the previous section and the basic concavity problems from earlier in this section. Because of that we will not be putting in quite as much detail here. If you are still unsure how to work the parts of this problem you should go back and work a few of the basic problems from the previous section and/or earlier in this section before proceeding. We will need the 1st derivative to start things off.

( ) ( )4 3 35 20 5 4g t t t t t′ = − = −

From the 1st derivative we can see that the critical points of this function are then, 0 & 4t t= =

(b) Determine the open intervals on which the function increases and decreases. To answer this part all we need is the number line for the 1st derivative.


( ) ( ) ( )Increasing : ,0 & 4, Decreasing : 0, 4−∞ ∞ (c) Classify the critical points as relative maximums, relative minimums or neither. From the number line in the previous step we get the following classifications of the critical points.


Calculus I


0 : Relative Maximum 4 : Relative Minimumt t= =

(d) Determine the open intervals on which the function is concave up and concave down. We’ll need the 2nd derivative to find the list of possible inflection points. ( ) ( )3 2 220 60 20 3g t t t t t′′ = − = − The possible inflection points for this problem are, 0 & 3t t= = To get the intervals of concavity we’ll need the number line for the 2nd derivative.

From this we get the following concavity information for the function.

( ) ( ) ( )Concave Up : 3, Concave Down : ,0 & 0,3∞ −∞ (e) Determine the inflection points of the function. From the concavity information in the previous step we can see that the single inflection point for the function is,

3t = (f) Use the information from steps (a) – (e) to sketch the graph of the function. Here is a sketch of the graph of this function using the information above. As we did in problems in this section we can start at the left and work our way to the right on the graph. As we do this we first pay attention to the increasing/decreasing information and then make sure that the curve has the correct concavity as we sketch it in.


Calculus I


Note that because we used a computer to generate the sketch it is possible that your sketch won’t be quite the same. It should however, have the same points listed on the graph above, the same basic increasing/decreasing nature and the same basic concavity. 10. For ( ) 3 45 8f x x x= − − answer each of the following questions.

(a) Identify the critical points of the function. (b) Determine the open intervals on which the function increases and decreases. (c) Classify the critical points as relative maximums, relative minimums or neither. (d) Determine the open intervals on which the function is concave up and concave down. (e) Determine the inflection points of the function. (f) Use the information from steps (a) – (e) to sketch the graph of the function.


( ) ( )2 3 224 4 4 6f x x x x x′ = − − = − +

From the 1st derivative we can see that the critical points of this function are then,


Calculus I


6 & 0x x= − =



( ) ( ) ( )Increasing : , 6 Decreasing : 6,0 & 0,−∞ − − ∞ (c) Classify the critical points as relative maximums, relative minimums or neither. From the number line in the previous step we get the following classifications of the critical points.

6 : Relative Maximum 0 : Neitherx x= − =

(d) Determine the open intervals on which the function is concave up and concave down. We’ll need the 2nd derivative to find the list of possible inflection points. ( ) ( )248 12 12 4f x x x x x′′ = − − = − + The possible inflection points for this function are, 4 & 0x x= − = To get the intervals of concavity we’ll need the number line for the 2nd derivative.


Calculus I



( ) ( ) ( )Concave Up : 4,0 Concave Down : , 4 & 0,− −∞ − ∞ (e) Determine the inflection points of the function. From the concavity information in the previous step we can see that the inflection points for the function are,

4 & 0x x= − = (f) Use the information from steps (a) – (e) to sketch the graph of the function. Here is a sketch of the graph of this function using the information above. As we did in problems in this section we can start at the left and work our way to the right on the graph. As we do this we first pay attention to the increasing/decreasing information and then make sure that the curve has the correct concavity as we sketch it in.

Note that because we used a computer to generate the sketch it is possible that your sketch won’t be quite the same. It should however, have the same points listed on the graph above, the same basic increasing/decreasing nature and the same basic concavity.


Calculus I


11. For ( ) 4 3 22 12h z z z z= − − answer each of the following questions.



( ) ( )3 2 24 6 24 2 2 3 12h z z z z z z z′ = − − = − −

From the 1st derivative we can see that the critical points of this function are then,

3 10540 & 1.8117, 3.3117x x ±= = = −




Calculus I


( ) ( ) ( ) ( )3 105 3 105 3 105 3 1054 4 4 4Increasing : ,0 & , Decreasing : , & 0,− + − +∞ −∞

(c) Classify the critical points as relative maximums, relative minimums or neither. From the number line in the previous step we get the following classifications of the critical points.

3 105

4

3 1054

: Relative Minimum 0 : Relative Maximum

: Relative Minimum

z z

z

−

+

= =

=

(d) Determine the open intervals on which the function is concave up and concave down. We’ll need the 2nd derivative to find the list of possible inflection points. ( ) ( ) ( )212 12 24 12 2 1h z z z z z′′ = − − = − + The possible inflection points for this function are, 1 & 2z z= − = To get the intervals of concavity we’ll need the number line for the 2nd derivative.


( ) ( ) ( )Concave Up : , 1 & 2, Concave Down : 1, 2−∞ − ∞ − (e) Determine the inflection points of the function. From the concavity information in the previous step we can see that the inflection points for the function are,


Calculus I


1 & 2z z= − = (f) Use the information from steps (a) – (e) to sketch the graph of the function. Here is a sketch of the graph of this function using the information above. As we did in problems in this section we can start at the left and work our way to the right on the graph. As we do this we first pay attention to the increasing/decreasing information and then make sure that the curve has the correct concavity as we sketch it in.

Note that because we used a computer to generate the sketch it is possible that your sketch won’t be quite the same. It should however, have the same points listed on the graph above, the same basic increasing/decreasing nature and the same basic concavity. 12. For ( ) ( )23 8sin tQ t t= − on [ ]7,4− answer each of the following questions.


(a) Identify the critical points of the function. The parts to this problem (with the exception of the last part) are just like the basic increasing/decreasing problems from the previous section and the basic concavity problems from


Calculus I


earlier in this section. Because of that we will not be putting in quite as much detail here. If you are still unsure how to work the parts of this problem you should go back and work a few of the basic problems from the previous section and/or earlier in this section before proceeding. We will need the 1st derivative to start things off.

( ) ( )23 4cos tQ t′ = −

From the 1st derivative all of the critical points are,

1.4454 4

0, 1, 2, 3,11.1210 4

t nn

t nπ

π= +

= ± ± ±= +

K

If you need some review of the solving trig equation process go back to the Solving Trig Equations sections for some examples. Plugging in some values of n we see that the critical points in the interval [ ]7,4− are,

1.4454 & 1.4454t t= − =



[ ) ( ] ( )Increasing : 7, 1.4454 & 1.4454, 4 Decreasing : 1.4454,1.4454− − − (c) Classify the critical points as relative maximums, relative minimums or neither. From the number line in the previous step we get the following classifications of the critical points.


Calculus I


1.4454 : Relative Maximum 1.4454 : Relative Minimumt t= − =

(d) Determine the open intervals on which the function is concave up and concave down. We’ll need the 2nd derivative to find the list of possible inflection points. ( ) ( )22sin tQ t′′ = All possible critical points of the function are,

4

0, 1, 2, 3,2 4

t nn

t nππ π

== ± ± ±

= +K

Plugging in some values of n we see that the possible inflection points in the interval [ ]7,4− are,

6.2832 & 0t t= − = To get the intervals of concavity we’ll need the number line for the 2nd derivative.


[ ) ( ] ( )Concave Up : 7, 6.2832 & 0,4 Concave Down : 6.2832,0− − − (e) Determine the inflection points of the function. From the concavity information in the previous step we can see that the inflection points for the function are,

6.2832 & 0t t= − = (f) Use the information from steps (a) – (e) to sketch the graph of the function. Here is a sketch of the graph of this function using the information above. As we did in problems in this section we can start at the left and work our way to the right on the graph. As we do this


Calculus I


we first pay attention to the increasing/decreasing information and then make sure that the curve has the correct concavity as we sketch it in.

Note that because we used a computer to generate the sketch it is possible that your sketch won’t be quite the same. It should however, have the same points listed on the graph above, the same basic increasing/decreasing nature and the same basic concavity.

13. For ( ) ( )43 2f x x x= − answer each of the following questions.




Calculus I


( ) ( ) ( )7 4 4 1 13 3 3 3 37 8 1

3 3 32 7 8f x x x f x x x x x′= − → = − = −

Note that by factoring the 13x out we made it a little easier to quickly see that the critical points

are, 8

70 & 1.1429x x= = =



( ) ( ) ( )8 87 7Increasing : ,0 & , Decreasing : 0,−∞ ∞

(c) Classify the critical points as relative maximums, relative minimums or neither. From the number line in the previous step we get the following classifications of the critical points.

870 : Relative Maximum : Relative Minimumx x= =

(d) Determine the open intervals on which the function is concave up and concave down. We’ll need the 2nd derivative to find the list of possible inflection points.

( )1 23 3

23

28 89 9

28 89xf x x xx

− −′′ = − =

The possible inflection points for this function are,


Calculus I


270 & 0.2857x x= = =

To get the intervals of concavity we’ll need the number line for the 2nd derivative.


( ) ( ) ( )2 27 7Concave Up : , Concave Down : ,0 & 0,∞ −∞

(e) Determine the inflection points of the function. From the concavity information in the previous step we can see that the single inflection point for the function is,

27x =

(f) Use the information from steps (a) – (e) to sketch the graph of the function. Here is a sketch of the graph of this function using the information above. As we did in problems in this section we can start at the left and work our way to the right on the graph. As we do this we first pay attention to the increasing/decreasing information and then make sure that the curve has the correct concavity as we sketch it in.


Calculus I


Note that because we used a computer to generate the sketch it is possible that your sketch won’t be quite the same. It should however, have the same points listed on the graph above, the same basic increasing/decreasing nature and the same basic concavity. 14. For ( ) 4wP w w= e on [ ]1

42,− answer each of the following questions.


(a) Identify the critical points of the function. The parts to this problem (with the exception of the last part) are just like the basic increasing/decreasing problems from the previous section and the basic concavity problems from earlier in this section. Because of that we will not be putting in quite as much detail here. If you are still unsure how to work the parts of this problem you should go back and work a few of the basic problems from the previous section and/or earlier in this section before proceeding. Also note that the interval here is only because we haven’t discussed L’Hospital’s Rule yet (that comes in a few sections…) and that makes the behavior of the graph as w → ± ∞ a little trickier. Once we cover that section you might want to come back and eliminate the interval and see what the graph is. We will need the 1st derivative to start things off.

( ) ( )4 4 44 1 4w w wP w w w′ = + = +e e e

From the 1st derivative we can see that the only critical points of this function is, 1

4w = −



Calculus I



( ) ( )1 14 4Increasing : , Decreasing : ,− ∞ −∞ −

(c) Classify the critical points as relative maximums, relative minimums or neither. From the number line in the previous step we get the following classification of the critical point.

14 : Relative Minimumw = −

(d) Determine the open intervals on which the function is concave up and concave down. We’ll need the 2nd derivative to find the list of possible inflection points. ( ) ( ) ( ) ( )4 4 44 4 1 4 8 1 2w w wP w w w′′ = + + = +e e e The only possible inflection point for this function is, 1

2w = − To get the intervals of concavity we’ll need the number line for the 2nd derivative.



Calculus I


( ) ( )1 12 2Concave Up : , Concave Down : ,− ∞ −∞ −

(e) Determine the inflection points of the function. From the concavity information in the previous step we can see that the single inflection point for the function is,

12w = −

(f) Use the information from steps (a) – (e) to sketch the graph of the function. Here is a sketch of the graph of this function using the information above. As we did in problems in this section we can start at the left and work our way to the right on the graph. As we do this we first pay attention to the increasing/decreasing information and then make sure that the curve has the correct concavity as we sketch it in.

Note that because we used a computer to generate the sketch it is possible that your sketch won’t be quite the same. It should however, have the same points listed on the graph above, the same basic increasing/decreasing nature and the same basic concavity. 15. Determine the minimum degree of a polynomial that has exactly one inflection point. Hint : What is the simplest possible form of the 2nd derivative that we can have that will guarantee that we have a single inflection point? Step 1


Calculus I


First, let’s suppose that the single inflection point occurs at x a= for some number a. The value of a is not important, this only allows us to discuss the problem. Now, if we start with a polynomial, call it ( )p x , then the 2nd derivative must also be a

polynomial and we have to have ( ) 0p a′′ = . In addition we know that the 2nd derivative must

change signs at x a= . The simplest polynomial that we can have that will do this is,

( )p x x a′′ = −

This clearly has ( ) 0p a′′ = and it will change sign at x a= . Note as well that we don’t really

care which side is concave up and which side is concave down. We only care that the 2nd derivative changes sign at x a= as it does here. Hint : We saw how to “undo” differentiation in the practice problems in the previous section. Here we simply need to do that twice and note that we don’t actually have to undo the derivatives here, just think about what they would have to look like. Step 2 Okay, saw how to “undo” differentiation in the practice problems of the previous section. We don’t actually need to do that here, but we do need to think about what undoing differentiation will give here. The 2nd derivative is a 1st degree polynomial and that means the 1st derivative had to be a 2nd degree polynomial. This should make sense to you if you understand how differentiation works. We know that we have to differentiate the 1st derivative to get the 2nd derivative. Therefore because the highest power of x in the 2nd derivative is 1 and we know that differentiation lowers the power by 1 the highest power of x in the 1st derivative must have been 2. Okay, we’ve figured out that the 1st derivative must have been a 2nd degree polynomial. This in turn means that the original function must have been a 3rd degree polynomial Again, differentiation lowers the power of x by 1 and if the highest power of x in the 1st derivative is 2 then the highest power of x in the original function must have been 3. So, the minimum degree of a polynomial that has exactly one inflection point must be three (i.e. a cubic polynomial).


Calculus I


Note that we can have higher degree polynomials with exactly one inflection point. This is simply the minimal degree that will give exactly one inflection point. 16. Suppose that we know that ( )f x is a polynomial with critical points 1x = − , 2x = and

6x = . If we also know that the 2nd derivative is ( ) 23 14 4f x x x′′ = + − . If possible, classify

each of the critical points as relative minimums, relative maximums. If it is not possible to classify the critical points clearly explain why they cannot be classified. Hint : We do NOT need the 1st derivative to answer this question. We are in the 2nd derivative section and we did see a way in the notes on how to use the 2nd derivative (which we have nicely been given…) to classify most critical points. Solution This problem is not as difficult as many students originally make it out to be. We’ve been given the 2nd derivative and we saw how the 2nd derivative test can be used to classify most critical points so let’s use that. First, we should note that because we have been told that ( )f x is a polynomial it should be

fairly clear that, regardless of what the 1st derivative actually is, we should have, ( ) ( ) ( )1 0 2 0 6 0f f f′ ′ ′− = = = What this means is that we can use the 2nd derivative test as it only works for these kinds of critical points. All we need to do then is plug the critical points into the 2nd derivative and use the 2nd derivative test to classify the critical points.

( )( )( )

1 21 0 Relative Maximum

2 12 0 Relative Minimum

6 28 0 Relative Maximum

f

f

f

′′ − = − <

′′ = >

′′ = − <

So, in this case it was possible to classify all of the given critical points. Recall that if the 2nd derivative had been zero for any of them we would not have been able to classify that critical point without the 1st derivative which we don’t have for this case.


Calculus I


The Mean Value Theorem 1. Determine all the number(s) c which satisfy the conclusion of Rolle’s Theorem for

( ) 2 2 8f x x x= − − on [ ]1,3− .

Solution The first thing we should do is actually verify that Rolle’s Theorem can be used here. The function is a polynomial which is continuous and differentiable everywhere and so will be continuous on [ ]1,3− and differentiable on ( )1,3− .

Next, a couple of quick function evaluations shows that ( ) ( )1 3 5f f− = = − .

Therefore, the conditions for Rolle’s Theorem are met and so we can actually do the problem. Note that this may seem to be a little silly to check the conditions but it is a really good idea to get into the habit of doing this stuff. Since we are in this section it is pretty clear that the conditions will be met or we wouldn’t be asking the problem. However, once we get out of this section and you want to use the Theorem the conditions may not be met. If you are in the habit of not checking you could inadvertently use the Theorem on a problem that can’t be used and then get an incorrect answer. Now that we know that Rolle’s Theorem can be used there really isn’t much to do. All we need to do is take the derivative, ( ) 2 2f x x′ = − and then solve ( ) 0f c′ = .

2 2 0 1c c− = ⇒ = So, we found a single value and it is in the interval so the value we want is,

1c =


Calculus I


2. Determine all the number(s) c which satisfy the conclusion of Rolle’s Theorem for

( ) 2 32g t t t t= − − on [ ]2,1− .

Solution The first thing we should do is actually verify that Rolle’s Theorem can be used here. The function is a polynomial which is continuous and differentiable everywhere and so will be continuous on [ ]2,1− and differentiable on ( )2,1− .

Next, a couple of quick function evaluations shows that ( ) ( )2 1 0g g− = = .

Therefore, the conditions for Rolle’s Theorem are met and so we can actually do the problem. Note that this may seem to be a little silly to check the conditions but it is a really good idea to get into the habit of doing this stuff. Since we are in this section it is pretty clear that the conditions will be met or we wouldn’t be asking the problem. However, once we get out of this section and you want to use the Theorem the conditions may not be met. If you are in the habit of not checking you could inadvertently use the Theorem on a problem that can’t be used and then get an incorrect answer. Now that we know that Rolle’s Theorem can be used there really isn’t much to do. All we need to do is take the derivative, ( ) 22 2 3g t t t′ = − − and then solve ( ) 0g c′ = .

2 1 733 2 2 0 1.2153, 0.5486c c c ±

−− − + = ⇒ = = − So, we found two values and, in this case, they are both in the interval so the values we want are,

1 73 1.2153, 0.5486c ±

−= = − 3. Determine all the number(s) c which satisfy the conclusion of Mean Value Theorem for

( ) 3 24 8 7 2h z z z z= − + − on [ ]2,5 .

Solution The first thing we should do is actually verify that the Mean Value Theorem can be used here.


Calculus I


The function is a polynomial which is continuous and differentiable everywhere and so will be continuous on [ ]2,5 and differentiable on ( )2,5 .

Therefore, the conditions for the Mean Value Theorem are met and so we can actually do the problem. Note that this may seem to be a little silly to check the conditions but it is a really good idea to get into the habit of doing this stuff. Since we are in this section it is pretty clear that the conditions will be met or we wouldn’t be asking the problem. However, once we get out of this section and you want to use the Theorem the conditions may not be met. If you are in the habit of not checking you could inadvertently use the Theorem on a problem that can’t be used and then get an incorrect answer. Now that we know that the Mean Value Theorem can be used there really isn’t much to do. All we need to do is do some function evaluations and take the derivative. ( ) ( ) ( ) 22 12 5 333 12 16 7h h h z z z′= = = − + The final step is to then plug into the formula from the Mean Value Theorem and solve for c. 2 2333 12

5 212 16 7 107 12 16 100 0c c c c−−− + = = → − − =

2 793 2.2961, 3.6294c ±= = −

So, we found two values and, in this case, only the second is in the interval and so the value we want is,

2 793 3.6294c += =

4. Determine all the number(s) c which satisfy the conclusion of Mean Value Theorem for

( ) 38 tA t t −= + e on [ ]2,3− .

Solution The first thing we should do is actually verify that the Mean Value Theorem can be used here. The function is a sum of a polynomial and an exponential function both of which are continuous and differentiable everywhere. This in turn means that the sum is also continuous and differentiable everywhere and so the function will be continuous on [ ]2,3− and differentiable on

( )2,3− .


Calculus I


Therefore, the conditions for the Mean Value Theorem are met and so we can actually do the problem. Note that this may seem to be a little silly to check the conditions but it is a really good idea to get into the habit of doing this stuff. Since we are in this section it is pretty clear that the conditions will be met or we wouldn’t be asking the problem. However, once we get out of this section and you want to use the Theorem the conditions may not be met. If you are in the habit of not checking you could inadvertently use the Theorem on a problem that can’t be used and then get an incorrect answer. Now that we know that the Mean Value Theorem can be used there really isn’t much to do. All we need to do is do some function evaluations and take the derivative. ( ) ( ) ( )6 9 32 16 3 24 8 3 tA A A t− −− = − + = + = −e e e The final step is to then plug into the formula from the Mean Value Theorem and solve for c.

( )( )

( )

9 624 1633 2

3

3

8 3 72.6857

3 80.685726.8952

3 ln 26.8952 3.29195 1.0973

c

c

c

c c

−+ − − +−− −

−

−

− = = −

=

=

− = = ⇒ = −

e ee

ee

So, we found a single value and it is in the interval and so the value we want is,

1.0973c = − 5. Suppose we know that ( )f x is continuous and differentiable on the interval [ ]7,0− , that

( )7 3f − = − and that ( ) 2f x′ ≤ . What is the largest possible value for ( )0f ?

Step 1 We were told in the problem statement that the function (whatever it is) satisfies the conditions of the Mean Value Theorem so let’s start out this that and plug in the known values. ( ) ( ) ( ) ( ) ( ) ( )0 7 7 0 0 3 7f f f c f f c′ ′− − = − → + = Step 2


Calculus I


Next, let’s solve for ( )0f .

( ) ( )0 7 3f f c′= − Step 3 Finally, let’s take care of what we know about the derivative. We are told that the maximum value of the derivative is 2. So, plugging the maximum possible value of the derivative into

( )f c′ above will, in this case, give us the maximum value of ( )0f . Doing this gives,

( ) ( ) ( )0 7 3 7 2 3 11f f c′= − ≤ − = So, the largest possible value for ( )0f is 11. Or, written as an inequality this would be written

as,

( )0 11f ≤ 6. Show that ( ) 3 27 25 8f x x x x= − + + has exactly one real root.

Hint : Can you use the Intermediate Value Theorem to prove that it has at least one real root? Step 1 First let’s note that ( )0 8f = . If we could find a function value that was negative the

Intermediate Value Theorem (which can be used here because the function is continuous everywhere) would tell us that the function would have to be zero somewhere. In other words, there would have to be at least one real root. Because the largest power of x is 3 it looks like if we let x be large enough and negative the function should also be negative. All we need to do is start plugging in negative x’s until we find one that works. In fact, we don’t even need to do much : ( )1 25f − = − .

So, we can see that ( ) ( )25 1 0 0 8f f− = − < < = and so by the Intermediate Value Theorem the

function must be zero somewhere in the interval ( )1,0− . The interval itself is not important.

What is important is that we have at least one real root. Hint : What would happen if there were more than one real root? Step 2


Calculus I


Next, let’s assume that there is more than one real root. Assuming this means that there must be two numbers, say a and b, so that, ( ) ( ) 0f a f b= = Next, because ( )f x is a polynomial it is continuous and differentiable everywhere and so we

could use Rolle’s Theorem to see that there must be a real value, c, so that, ( ) 0f c′ = Note that Rolle’s Theorem tells us that c must be between a and b. Since both of these are real values then c must also be real. Hint : Is that possible? Step 3 Because ( )f x is a polynomial it is easy enough to see if such a c exists.

( ) 2 2 7 263 14 25 3 14 25 03

if x x x c c c ±′ = − + → − + = → =

So, we can see that in fact the only two places where the derivative is zero are complex numbers are not real numbers. Therefore, it is not possible for there to be more than one real root. From Step 1 we know that there is at least one real root and we’ve just proven that we can’t have more than one real root. Therefore, there must be exactly one real root.

Optimization 1. Find two positive numbers whose sum is 300 and whose product is a maximum. Step 1 The first step is to write down equations describing this situation. Let’s call the two numbers x and y and we are told that the sum is 300 (this is the constraint for the problem) or,


Calculus I


300x y+ = We are being asked to maximize the product, A xy= Step 2 We now need to solve the constraint for x or y (and it really doesn’t matter which variable we solve for in this case) and plug this into the product equation. ( ) ( ) 2300 300 300y x A x x x x x= − ⇒ = − = − Step 3 The next step is to determine the critical points for this equation. ( ) 300 2 300 2 0 150A x x x x′ = − → − = → = Step 4 Now for the step many neglect as unnecessary. Just because we got a single value we can’t just assume that this will give a maximum product. We need to do a quick check to see if it does give a maximum. As discussed in notes there are several methods for doing this, but in this case we can quickly see that, ( ) 2A x′′ = − From this we can see that the second derivative is always negative and so ( )A x will always be

concave down and so the single critical point we got in Step 3 must be a relative maximum and hence must be the value that gives a maximum product. Step 5 Finally, let’s actually answer the question. We need to give both values. We already have x so we need to determine y and that is easy to do from the constraint. 300 150 150y = − = The final answer is then,

150 150x y= =


Calculus I


2. Find two positive numbers whose product is 750 and for which the sum of one and 10 times the other is a minimum. Step 1 The first step is to write down equations describing this situation. Let’s call the two numbers x and y and we are told that the product is 750 (this is the constraint for the problem) or,

750xy = We are then being asked to minimize the sum of one and 10 times the other, 10S x y= + Note that it really doesn’t worry which is x and which is y in the sum so we simply chose the y to be multiplied by 10. Step 2 We now need to solve the constraint for x or y (and it really doesn’t matter which variable we solve for in this case) and plug this into the product equation.

( )750 750 10x S y yy y

= ⇒ = +

Step 3 The next step is to determine the critical points for this equation.

( ) 2 2

750 75010 10 0 75 5 3S y yy y

′ = − + → − + = → = ± = Because we are told that y must be positive we can eliminate the negative value and so the only

value we really get out of this step is : 75 5 3y = = . Step 4 Now for the step many neglect as unnecessary. Just because we got a single value we can’t just assume that this will give a minimum sum. We need to do a quick check to see if it does give a minimum. As discussed in notes there are several methods for doing this, but in this case we can quickly see that,


Calculus I


( ) 3

1500S yy

′′ =

From this we can see that, provided we recall that y is positive, then the second derivative will always be positive. Therefore, ( )S y will always be concave up and so the single critical point

from Step 3 that we can use must be a relative minimum and hence must be the value that gives a minimum sum. Step 5 Finally, let’s actually answer the question. We need to give both values. We already have y so we need to determine x and that is easy to do from the constraint.

750 50 35 3

x = =

The final answer is then,

50 3 5 3x y= = 3. Let x and y be two positive numbers such that 2 50x y+ = and ( ) ( )1 2x y+ + is a maximum.

Step 1 In this case we were given the constraint in the problem, 2 50x y+ = We are also told the equation to maximize, ( )( )1 2f x y= + + So, let’s just solve the constraint for x or y (we’ll solve for x to avoid fractions…) and plug this into the product equation. ( ) ( )( ) ( ) ( ) 250 2 50 2 1 2 51 2 2 102 47 2x y f y y y y y y y= − ⇒ = − + + = − + = + − Step 2 The next step is to determine the critical points for this equation.

( ) 4747 4 47 4 04

f y y y y′ = − → − = → =


Calculus I


Step 3 Now for the step many neglect as unnecessary. Just because we got a single value we can’t just assume that this will give a maximum product. We need to do a quick check to see if it does give a maximum. As discussed in notes there are several methods for doing this, but in this case we can quickly see that, ( ) 4f y′′ = − From this we can see that the second derivative is always negative and so ( )f y will always be

concave down and so the single critical point we got in Step 2 must be a relative maximum and hence must be the value that gives a maximum. Step 4 Finally, let’s actually answer the question. We need to give both values. We already have y so we need to determine x and that is easy to do from the constraint.

47 5350 24 2

x = − =

The final answer is then,

53 472 4

x y= =

4. We are going to fence in a rectangular field. If we look at the field from above the cost of the vertical sides are $10/ft, the cost of the bottom is $2/ft and the cost of the top is $7/ft. If we have $700 determine the dimensions of the field that will maximize the enclosed area. Step 1 The first step is to do a quick sketch of the problem. We could probably skip the sketch in this case, but that is a really bad habit to get into. For many of these problems a sketch is really convenient and it can be used to help us keep track of some of the important information in the problem and to “define” variables for the problem. Here is the sketch for this problem.


Calculus I


Step 2 Next we need to set up the constraint and equation that we are being asked to optimize. We are told that we have $700 to spend and so the cost of the material will be the constraint for this problem. The cost for the material is then, 700 10 2 10 7 20 9y x y x y x= + + + = + We are being asked to maximize the area so that equation is, A xy= Step 3 Now, let’s solve the constraint for y (that looks like it will only have one fraction in it and so may be “easier”…). 9

2035y x= − Plugging this into the area formula gives, ( ) ( ) 29 9

20 2035 35A x x x x x= − = − Step 4 Finding the critical point(s) for this shouldn’t be too difficult at this point so here is that work. ( ) 9 9 350

10 10 935 35 0A x x x x′ = − → − = → = Step 5 The second derivative of the area function is, ( ) 9

10A x′′ = −


Calculus I


From this we can see that the second derivative is always negative and so ( )A x will always be

concave down and so the single critical point we got in Step 4 must be a relative maximum and hence must be the value that gives a maximum area. Step 6 Now, let’s finish the problem by getting the second dimension. ( )9 350 35

20 9 235y = − = The final dimensions are then,

350 359 2x y= =

5. We have 45 m2 of material to build a box with a square base and no top. Determine the dimensions of the box that will maximize the enclosed volume. Step 1 The first step is to do a quick sketch of the problem. We could probably skip the sketch in this case, but that is a really bad habit to get into. For many of these problems a sketch is really convenient and it can be used to help us keep track of some of the important information in the problem and to “define” variables for the problem. Here is the sketch for this problem.

Step 2 Next we need to set up the constraint and equation that we are being asked to optimize. We are told that we have 45 m2 of material to build the box and so that is the constraint. The amount of material that we need to build the box is then, ( ) ( ) 2 245 2 2 2 2 4lw lh wh w wh wh w wh= + + = + + = +


Calculus I


Note that because there is no top the first term won’t have the 2 that the second and third term have. Be careful with this kind of thing it is easy to miss if you aren’t paying attention. We are being asked to maximize the volume so that equation is, 2V lwh w h= = Note as well that we went ahead and used fact that l w= in both of these equations to reduce the three variables in the equation down to two variables. Step 3 Now, let’s solve the constraint for h (that will allow us to avoid dealing with roots, plus there is only one h in the constraint so it will simply be easier to deal with).

245

4wh

w−

=

Plugging this into the volume formula gives,

( ) ( ) ( )2

2 2 345 1 145 454 4 4

wV w w w w w ww

−= = − = −

Step 4 Finding the critical point(s) for this shouldn’t be too difficult at this point so here is that work.

( ) ( ) ( )2 2 451 14 4 345 3 45 3 0 15V w w w w′ = − → − = → = ± = ±

Because we are dealing with the dimensions of a box the negative width doesn’t make any sense

and so the only critical point that we can use here is : 15w = . Be careful here and do not get into the habit of just eliminating the negative values. The only reason for eliminating it in this case is for physical reasons. If we had just given the equations without any physical reasoning it would have to be included in the rest of the work! Step 5 The second derivative of the volume function is, ( ) 6V w w′′ = − From this we can see that the second derivative is always negative for positive w (which we will always have for this case since w is the width of a box). Therefore, provided w is positive,


Calculus I


( )V w will always be concave down and so the single critical point we got in Step 4 must be a

relative maximum and hence must be the value that gives a maximum volume. Step 6 Now, let’s finish the problem by getting the remaining dimensions.

45 1515 3.8730 1.93654 15

l w h −= = = = =

The final dimensions are then,

3.8730 1.9365l w h= = = 6. We want to build a box whose base length is 6 times the base width and the box will enclose 20 in3. The cost of the material of the sides is $3/in2 and the cost of the top and bottom is $15/in2. Determine the dimensions of the box that will minimize the cost. Step 1 The first step is to do a quick sketch of the problem. We could probably skip the sketch in this case, but that is a really bad habit to get into. For many of these problems a sketch is really convenient and it can be used to help us keep track of some of the important information in the problem and to “define” variables for the problem. Here is the sketch for this problem.

Step 2 Next we need to set up the constraint and equation that we are being asked to optimize. We are told that the volume of the box must be 20 in3 and so this is the constraint. 220 6lwh w h= =


Calculus I


We are being asked to minimize the cost and the cost function is,

( ) ( ) ( ) [ ] 2 23 2 2 15 2 3 12 2 15 12 42 180C lh wh lw wh wh w wh w = + + = + + = + Note as well that we went ahead and used fact that 6l w= in both of these equations to reduce the three variables in the equation down to two variables. Step 3 Now, let’s solve the constraint for h (that will allow us to avoid dealing with roots).

2

103

hw

=

Plugging this into the cost function gives,

( ) 2 22

10 14042 180 1803

C w w w ww w

= + = +

Step 4 Finding the critical point(s) for this shouldn’t be too difficult at this point. Here is the derivative.

( )3

2 2

140 360 140360 wC w ww w

−′ = − + =

From this it looks like the critical points are : 0w = and 7318 0.7299w = = .

Because we are dealing with the dimensions of a box the zero width doesn’t make any sense and

so the only critical point that we can use here is : 7318 0.7299w = = .

Be careful here and do not get into the habit of just eliminating the zero as a critical point. The only reason for eliminating it in this case is for physical reasons. If we had just given the equations without any physical reasoning it would have to be included in the rest of the work! Step 5 The second derivative of the volume function is,

( ) 3

280 360C ww

′′ = +

From this we can see that the second derivative is always positive for positive w (which we will always have for this case since w is the width of a box). Therefore, provided w is positive,


Calculus I


( )C w will always be concave up and so the single critical point we got in Step 4 must be a

relative minimum and hence must be the value that gives a minimum cost. Step 6 Now, let’s finish the problem by getting the remaining dimensions.

( )2

106 4.3794 6.25683 0.7299

l w h= = = =


0.7299 4.3794 6.2568w l h= = = 7. We want to construct a cylindrical can with a bottom but no top that will have a volume of 30 cm3. Determine the dimensions of the can that will minimize the amount of material needed to construct the can. Step 1 The first step is to do a quick sketch of the problem. We could probably skip the sketch in this case, but that is a really bad habit to get into. For many of these problems a sketch is really convenient and it can be used to help us keep track of some of the important information in the problem and to “define” variables for the problem. Here is the sketch for this problem.

Step 2 Next we need to set up the constraint and equation that we are being asked to optimize. We are told that the volume of the can must be 30 cm3 and so this is the constraint. 230 r hπ=


Calculus I


We are being asked to minimize the amount of material needed to construct the can, 22A rh rπ π= + Recall that the can will have no top and so the second term will only be for the area of the bottom of the can. Step 3 Now, let’s solve the constraint for h (that will allow us to avoid dealing with roots).

2

30hrπ

=

Plugging this into the amount of material function gives,

( ) 2 22

30 602A r r r rr r

π π ππ

= + = +

Step 4 Finding the critical point(s) for this shouldn’t be too difficult at this point. Here is the derivative.

( )3

2 2

60 2 602 rA r rr r

ππ

−′ = − + =

From this it looks like the critical points are : 0r = and 6032 2.1216r π= = .

Because we are dealing with the dimensions of a can the zero radius doesn’t make any sense and

so the only critical point that we can use here is : 6032 2.1216r π= = .

Be careful here and do not get into the habit of just eliminating the zero as a critical point. The only reason for eliminating it in this case is for physical reasons. If we had just given the equations without any physical reasoning it would have to be included in the rest of the work! Step 5 The second derivative of the volume function is,

( ) 3

120 2A rr

π′′ = +

From this we can see that the second derivative is always positive for positive r (which we will always have for this case since r is the radius of a can). Therefore, provided r is positive, ( )A r


Calculus I


will always be concave up and so the single critical point we got in Step 4 must be a relative minimum and hence must be the value that gives a minimum amount of material. Step 6 Now, let’s finish the problem by getting the height of the can.

( )2

30 2.12152.1216

hπ

= =


2.1216 2.1215r h= = 8. We have a piece of cardboard that is 50 cm by 20 cm and we are going to cut out the corners and fold up the sides to form a box. Determine the height of the box that will give a maximum volume. Step 1 The first step is to do a quick sketch of the problem.

Step 2 As with the problem like this in the notes the constraint is really the size of the box and that has been taken into account in the figure so all we need to do is set up the volume equation that we want to maximize. ( ) ( )( ) 3 250 2 20 2 4 140 1000V h h h h h h h= − − = − + Step 3 Finding the critical point(s) for this shouldn’t be too difficult at this point so here is that work,

( ) 2 35 5 19312 280 1000 4.4018, 18.9315V h h h h ±′ = − + = =


Calculus I


From the figure above we can see that the limits on h must be 0h = and 10h = (the largest h could is ½ the smaller side). Note that neither of these really make physical sense but they do provide limits on h. So, we must have 0 10h≤ ≤ and this eliminates the second critical point and so the only critical point we need to worry about is 4.4018h = Step 4 Because we have limits on h we can quickly check to see if we have maximum by plugging in the volume function. ( ) ( ) ( )0 0 4.4018 2030.34 10 0V V V= = =

So, we can see then that the height of the box will have to be 4.4018h = in order to get a maximum volume.

More Optimization Problems 1. We want to construct a window whose middle is a rectangle and the top and bottom of the window are semi-circles. If we have 50 meters of framing material what are the dimensions of the window that will let in the most light? Step 1 Let’s start with a quick sketch of the window.


Calculus I


Step 2 Next we need to set up the constraint and equation that we are being asked to optimize. We are told that we have 50 meters of framing material (i.e. the perimeter of the window) and so that will be the constraint for this problem. ( )50 2 2 2 2h r h rπ π= + = + We are being asked to maximize the amount of light being let in and that is simply the enclosed area or, ( ) ( )2 21

22 2 2A h r r hr rπ π= + = + With both of these equations we were a little careful with the last term. In each case we needed either the perimeter or area of each semicircle and there were two of them. The end result of course is the equation of the perimeter/area of a whole circle, but we really should be careful setting these equations up and note just where everything is coming from. Step 3 Now, let’s solve the constraint for h. 25h rπ= − Plugging this into the area function gives, ( ) ( ) 2 22 25 50A r r r r r rπ π π= − + = − Step 4 Finding the critical point(s) for this shouldn’t be too difficult at this point. Here is the derivative. ( ) 50 2A r rπ′ = − From this it looks like we get a single critical points are : 25 7.9577r π= = . Step 5 The second derivative of the volume function is, ( ) 2A r π′′ = − From this we can see that the second derivative is always negative. Therefore ( )A r will always

be concave down and so the single critical point we got in Step 4 must be a relative maximum and hence must be the value that allows in the maximum amount of light. Step 6


Calculus I


Now, let’s finish the problem by getting the radius of the semicircles. ( )2525 0h ππ= − = Okay, what this means is that in fact the most light will come from not even having a rectangle between the semicircles and just having a circular window of radius 25r π= . 2. Determine the area of the largest rectangle that can be inscribed in a circle of radius 1. Step 1 Let’s start with a quick sketch of the circle and rectangle. Also, in order to make the work a little easier we went ahead and assumed that the circle was centered at the origin of the standard xy-coordinate system. We’ve also defined a point ( ),x y in the first quadrant. This is the point that we will be

attempting to find when we get into the problems. If we know the coordinates of this point then the rectangle defined by the point, as shown in the figure, will be the one with the largest area.

Step 2 Next we need to set up the constraint and equation that we are being asked to optimize. Given our graph above we can easily determine the equation of the circle. This will also be the constraint of the problem because the corners of the rectangle must be on the circle. 2 2 1x y+ =


Calculus I


Also note that from the figure or equation we can clearly see that 1 1x− ≤ ≤ and 1 1y− ≤ ≤ . One or both of these limits will be useful later on in the problem. We are being asked to maximize the amount of the rectangle and using the definitions we see in the figure above the area is, ( )( )2 2 4A x y xy= = Step 3 We can solve the constraint for x or y. Either will lead to essentially the same work so we’ll solve for x. 21x y= ± − Because we’ve defined the point on the circle to be in the 1st quadrant we will use the “+” portion of this. Plugging this into the area function gives,

( ) 24 1A y y y= − Step 4 Finding the critical point(s) for this shouldn’t be too difficult at this point. Here is the derivative.

( )2 2

2

2 2

4 4 84 11 1

y yA y yy y

−′ = − − =− −

From this it looks like, from the numerator, we get the critical points,

1 12 2

0.7071y = ± = ± = ±

From the denominator we get the critical points : 1y = ± . Before proceeding to the next step let’s notice that because our point is in the first quadrant we know that y must be positive. This fact along with the limits on y we discussed in Step 2 tells us that we must have : 0 1y≤ ≤ . This in turn tells us that the only two critical points that we need to worry about are,

12

0.7071 1y y= = =

Step 5 Because we’ve got a range for possible critical points all we need to do to determine the maximum area is plug the end points and critical points into the area.


Calculus I


( ) ( ) ( )12

0 2 1 0A A A= = Step 6 So, the area of the largest rectangle that can be inscribed in the circle is : 2. 3. Find the point(s) on 23 2x y= − that are closest to ( )4,0− .

Step 1 Let’s start with a quick sketch of this situation. Below is a sketch of the graph of the function as well as the point ( )4,0− . As we can see we can expect to get two points as answers with the

only difference being the sign on the y-coordinate.

Step 2 Next we need to set up the constraint and equation that we are being asked to optimize. In this case the constraint is simply the equation we are given. The point must lie on the graph and so must also satisfy the equation. 23 2x y= − We are being asked to minimize the distance between a point (or points) on the graph and the point ( )4,0− . We can do this by looking at the distance between ( )4,0− and ( ),x y . The distance between these two points is,

( )2 24d x y= + +


Calculus I


As we discussed in the notes for this section the point that minimizes the square of the distance will also minimize the distance itself and so to avoid dealing with the root we will minimize the square of the distance or,

( )22 24d x y= + + Step 3 Now we have two choices on how to proceed from this point. The first option is to plug the equation we are given into the x in the distance squared and get a 4th degree polynomial for y that we’ll need to work with. The second is to solve the equation for

2y and plug that into the distance squared and get a 2nd degree polynomial for x that we’ll need to work with. The second option gives a “nicer” polynomial to work with so we’ll do that. ( )2 31 1

2 2 23y x x= − = − Plugging this into the distance squared gives,

( ) ( )22 23 15 3512 2 2 24f x d x x x x= = + + − = + +

Step 4 Finding the critical point(s) for this shouldn’t be too difficult at this point. Here is the derivative. ( ) 15

22f x x′ = + From this it looks like we get a single critical point : 15

4 3.75x = − = − . Step 5 The second derivative of the distance squared function is, ( ) 2f x′′ = From this we can see that the second derivative is always positive. Therefore the distance squared will always be concave up and so the single critical point we got in Step 4 must be a relative minimum and hence must be the value of x that gives the points that are closest to

( )4,0− .

Step 6 Finally we just need to determine the values y that give the actual points.


Calculus I


( )2 3 15 27 2712 2 4 8 8 1.8371y y= − − = ⇒ = ± = ±

So, the two points on the graph that are closest to ( )4,0− are,

( ) ( )15 27 15 274 4 4 4, & ,− − −

4. An 80 cm piece of wire is cut into two pieces. One piece is bent into an equilateral triangle and the other will be bent into a rectangle with one side 4 times the length of the other side. Determine where, if anywhere, the wire should be cut to maximize the area enclosed by the two figures. Step 1 Before we do a sketch we’ll need to do a little setup. Let’s suppose that the length of the piece of wire that goes to the rectangle is x. This means that the length of the piece of wire going to the triangle is 80 x− . We know that the length of each side of the triangle are equal and so must have length ( )1

3 80 x−

. We also know that the interior angles of the triangle are 3π and so the height of the triangle is

( ) ( ) ( )313 3 680 sin 80x xπ− = − .

For the rectangle let’s suppose that the length of the smaller side is L and so the length of the larger side is 4L. Next, we know that the total perimeter of the rectangle is x and so we must have, ( ) ( ) 102 2 4 10 xx L L L L= + = → = Now that we have all the various lengths of the figures in terms of x (which will make the work here a little easier) let’s summarize everything up with the following figure.


Calculus I


Step 2 Next we need to set up the constraint and equation that we are being asked to optimize. This is one of those cases where we really don’t have a constraint equation to work with. The constraint is the length of the wire (80 cm), but we took that into account when we set up our figure above so there isn’t anything to do with that in this case. We are being asked to maximize the enclosed area of the two figures and so here is the total area of the enclosed figures.

( ) ( ) ( ) ( ) ( ) ( )2 23 32 1 110 5 2 3 6 25 3680 80 80x x xA x x x x = + − − = + −

Step 3 Finding the critical point(s) for this shouldn’t be too difficult at this point (although the Algebra will be a little messy). Here is the derivative.

( ) ( )3225 18 80xA x x′ = − −

From this it looks like we get a single critical point,

40 3

932

25 18

43.6828x = =+

Step 4 The second derivative of the area function is,

( ) 3225 18A x′′ = +

From this we can see that the second derivative is always positive. Therefore ( )A x will always

be concave up and so the single critical point we got in Step 4 must be a relative minimum and hence must be the value of x (i.e. the cut point) that will give the minimum enclosed area. This is a problem however as we were asked for the maximum enclosed area. This is the reason for this step being in every problem that we’ve worked over the last couple of sections. Far too often students get to this point, get a single answer and then just assume that it must be the correct answer and don’t bother doing any kind of checking to verify if it is the correct answer. After all there was a single value so there is no choice for it to be correct. Right? Well, no. As we’ll seen here it in fact is not the correct answer. Step 5


Calculus I


So, what to do? We’ll recall for the problem statement that we were asked to, “Determine where, if anywhere, the wire should be cut to maximize the area enclosed by the two figures.” The “if anywhere” portion seems to suggest that we may not want to cut it at all. Maybe all of the wire should go to the rectangle (corresponding to 80x = above) or maybe all of the wire should to the triangle (corresponding to 0x = above). So, all we need to do is plug 80x = and 0x = into the area function and determine which will give the largest area.

( )

( )( )

0 307.92 All wire goes to triangle.

43.6828 139.785 Wire goes to both triangle and rectangle.

80 256 All wire goes to rectangle.

A

A

A

=

=

=

Note that we included the critical point above just to make it really clear that it will not in fact give the maximum area. We didn’t really need to include it here as we already knew it wouldn’t work for us. From the function evaluations above it looks like we’ll need to take all of the wire and bend it into an equilateral triangle in order to get the maximum area. 5. A line through the point ( )2,5 forms a right triangle with the x-axis and y-axis in the 1st

quadrant. Determine the equation of the line that will minimize the area of this triangle. Step 1 This problem may seem a little tricky at first. Here is a sketch of a line that goes through the point ( )2,5 , has an x-intercept of ( ),0a and a y-intercept of ( )0,b .


Calculus I


Note that the only way we can get a triangle with the line, x-axis and y-axis as sides is to require that 2a > and 5b > . If either of those are not true we will not have the triangle that we want. Step 2 Next we need to set up the constraint and equation that we are being asked to optimize. We are being asked to minimize the area of the triangle shown above. In terms of the quantities given on the graph it is easy enough to get an equation for the area. The base length of the triangle is a and the height of the triangle is b. We don’t have values for either of these but that isn’t a problem. Here is the area of the triangle. 1

2A ab= The constraint in this case is the equation of the line since that will define the hypotenuse of the triangle and hence also give both the base and height of the triangle. We need to write down the equation of the line, but we have three points on the line that we can use. Note however, that we really should use ( )2,5 as one of the points because the line does need to go the point and using

this point to write down the equation will give us that without any extra work. The real question then is whether we should use the x or y-intercept for the second point when determining the slope of the line. It really doesn’t matter which point that you use. The work will be slightly different for each point but there will be no real difference in the difficulty of the problem. We are going to use ( ),0a for the second point. The slope of the line using this point is,

52

ma

=−

We already know that b is the y-intercept and so the equation of the line through the point is,


Calculus I


52

y x ba

= +−

Note that we definitely seem to have a problem here. Normally at this point we’ve got two equation and two unknowns. In this case we appear to have four unknowns : a, b, x and y. This isn’t a problem as well see in the next step. Step 3 We now need to solve the constraint for one of the unknowns in the area function, i.e either a or b. However, as we noted above we also have an x and y in the equation that will cause problems if they stay in the equation. The point of this step is to get the area function down to a single variable. If we leave the x and y in the equation of the line we will end up with an area function with not one variable but three and that won’t work for us. What we really need is an equation involving only a and b that we can solve for one or the other and plug into the area function. Luckily this is easy to get. All we need to do is plug the x-intercept into the equation of the line to get,

502

a ba

= +−

Do you see why we couldn’t have used the y-intercept here? If not, plug it in and you’ll very quickly see why it won’t work. At this point we can easily solve the equation for b to get,

5 52 2

a aba a

= − =− −

To eliminate one of the minus signs we took the minus sign in front of the quotient and applied it to the denominator and simplified. This doesn’t need to be done, but it does eliminate one of them. Note that if we had used the y-intercept to determine the slope we would have found it to be easier at this step to solve for a instead. That is the only real difference in which point you use to find the slope. Okay, let’s put all this together. We know the value of b in terms of a so plug that into the area function to get,

( ) ( )21 5 5

2 2 2 2a aA a a

a a = = − −


Calculus I


Step 4 Here is the derivative of the area function,

( )( )

( )( )

2

2 2

45 4 52 22 2

a aa aA aa a

−−′ = =− −

From this it looks like we get a three possible critical points : 0a = , 2a = and 4a = . We can’t use 0a = as the critical point because that will no longer form a triangle with both the x-axis and the y-axis as the problem asks for as noted in the first step. We also can’t use 2a = for two reasons. First, it isn’t actually a critical point because the area function doesn’t exist at 2a = . This shouldn’t be surprising given that if we used this point we wouldn’t have a triangle anyway (again as we noted in the first step) and that is also the second reason for not using it. This leaves only 4a = as a potential critical point that we can use. Step 5 The second derivative of the area function (after a little simplification) is,

( )( )3

202

A aa

′′ =−

From this we can see that the second derivative is always positive provided we have 2a > . However, as we noted in the first step this is required in order even work the problem. Therefore, the second derivative will always be positive for the range of a that we are working on. The area function will then will always be concave up for the range of a and 4a = must give a minimum area. Step 6 Now that we know the value of a we know that the slope and y-intercept are,

( )5 45 5 102 4 2 4 2

m b= = − = =− −

The equation of the line is then,

5 102

y x= − +


Calculus I


6. A piece of pipe is being carried down a hallway that is 18 feet wide. At the end of the hallway there is a right-angled turn and the hallway narrows down to 12 feet wide. What is the longest pipe (always keeping it horizontal) that can be carried around the turn in the hallway? Step 1 Let’s start with a quick sketch of the pipe and hallways with all the important quantities given.

Step 2 Next we need to set up the constraint and equation that we are being asked to optimize. As we discussed in the similar problem in the notes for this section we actually need to minimize the total length of the pipe. The equation we need to minimize is then, 1 2L L L= + Also as we discussed in the notes problem with actually have two constraints : the widths of the two hallways. We can easily solve for these in terms of the angle θ . 1 212sec 18cscL Lθ θ= = As discussed in the notes problem we also know that we must have 20 πθ< < . Step 3 All we need to do here is plug our two constraints in the length function to get a function in terms of θ that we can minimize. ( ) 12sec 18cscL θ θ θ= + Step 4


Calculus I


The derivative of the length function is, ( ) 12sec tan 18csc cotL θ θ θ θ θ′ = − Next we need to set this equal to zero and solve this for θ to get the critical point that is in the range 20 πθ< < .

3

12sec tan 18csc cotsec tan 18csc cot 12

3tan2

θ θ θ θθ θθ θ

θ

=

=

=

The critical point that we need is then : ( )1 332tan 0.8528θ −= = .

Step 5 Verifying that this is the value that gives the minimum is a little trickier than the other problems. As noted in the notes for this section as we move 0θ → we have L → ∞ and as we move

2πθ → we have L → ∞ . Therefore, on either side of 0.8528θ = radians the length of the pipe

is increasing to infinity as we move towards the end of the range. Therefore, this angle must give us the minimum length of the pipe and so is the largest pipe that we can fit around corner. Step 6 The largest pipe that we can fit around the corner is then,

( )0.8528 42.1409 feetL = 7. Two 10 meter tall poles are 30 meters apart. A length of wire is attached to the top of each pole and it is staked to the ground somewhere between the two poles. Where should the wire be staked so that the minimum amount of wire is used? Step 1 Let’s start with a quick of the situation.


Calculus I


Step 2 Next we need to set up the constraint and equation that we are being asked to optimize. We want to minimize the amount of wire and so the equation we need to minimize is, 1 2L L L= + The constraint here is that the poles must be 30 meters apart. We can use this to determine the lengths of the individual wires in terms of x. Doing this gives,

( )221 2100 100 30L x L x= + = + −

Note as well that can also see that we need to require that 0 30x≤ ≤ . Step 3 All we need to do here is plug the lengths of the individual wires in the total length to get a function in terms of x that we can minimize.

( ) ( )22100 100 30L x x x= + + + − Step 4 The derivative of the length function is,

( )2 2

30100 60 1000

x xL xx x x

−′ = ++ − +

Solving for the critical point(s) is going to be messy so here it goes.


Calculus I


( )( ) ( ) ( )

2 2

2 2

2 2

22 2 2

4 3 2 4 3 2

30 0100 60 1000

30100 60 1000

60 1000 30 100

60 1000 30 100

60 1000 60 1000 6000 900000 6000 90000

15

x xx x x

x xx x x

x x x x x

x x x x x

x x x x x x xx

x

−+ =

+ − +−

= −+ − +

− + = − − +

− + = − +

− + = − + − += − +=

A quick check by plugging this back into the derivative shows that we do indeed get ( )15 0L′ =

and so this is a critical point and it is in the acceptable range of x. Recall that because we squared both sides of the equation above it is possible to end up with answers that in fact are not solutions and so we have to go back and check in the original equation to make sure that they are solutions. Step 5 Since we have a range of x’s and the distance function is continuous in the range all we need to do is plug in the endpoints and the critical point to identify the minimum distance. ( ) ( ) ( )0 41.6228 15 36.0555 30 41.6228L L L= = = Step 6 The wire should be staked midway between the poles to minimize the amount of wire.

Indeterminate Forms and L’Hospital’s Rule

1. Use L’Hospital’s Rule to evaluate 3 2

22

7 10lim6x

x x xx x→

− ++ −

.

Step 1 The first step we should really do here is verify that L’Hospital’s Rule can in fact be used on this limit.


Calculus I


This may seem like a silly step given that we are told to use L’Hospital’s Rule. However, in later sections we won’t be told to use it when/if it can be used. Therefore, we really need to get in the habit of checking that it can be used before applying it just to make sure that we can. If we apply L’Hospital’s Rule to a problem that it can’t be applied to then it’s is almost assured that we will get the wrong answer (it’s always possible you might get lucky and get the correct answer, but we will only be very lucky if it does). So a quick check shows us that,

3 2

2

7 10 0as 26 0

x x xxx x− +

→ →+ −

and so this is a form that allows the use of L’Hospital’s Rule. Step 2 So, at this point let’s just apply L’Hospital’s Rule.

3 2 2

22 2

7 10 3 14 10lim lim6 2 1x x

x x x x xx x x→ →

− + − +=

+ − +

Step 3 At this point all we need to do is try the limit and see if it can be done.

3 2 2

22 2

7 10 3 14 10 6lim lim6 2 1 5x x

x x x x xx x x→ →

− + − + −= =

+ − + So, the limit can be done and we done with the problem! The limit is then,

3 2

22

7 10 6lim6 5x

x x xx x→

− += −

+ −

2. Use L’Hospital’s Rule to evaluate ( )

24

sinlim

16w

ww

π→− −

.

Step 1 The first step we should really do here is verify that L’Hospital’s Rule can in fact be used on this limit. This may seem like a silly step given that we are told to use L’Hospital’s Rule. However, in later sections we won’t be told to use it when/if it can be used. Therefore, we really need to get in the habit of checking that it can be used before applying it just to make sure that we can. If we apply L’Hospital’s Rule to a problem that it can’t be applied to then it’s is almost assured that we will


Calculus I


get the wrong answer (it’s always possible you might get lucky and get the correct answer, but we will only be very lucky if it does). So a quick check shows us that,

( )2

sin 0as 416 0w

ww

π→ − →

− and so this is a form that allows the use of L’Hospital’s Rule. Step 2 So, at this point let’s just apply L’Hospital’s Rule.

( ) ( )

24 4

sin coslim lim

16 2w w

w ww w

π π π→− →−

=−

Step 3 At this point all we need to do is try the limit and see if it can be done.

( ) ( ) ( )24 4

sin cos cos 4lim lim

16 2 8 8w w

w ww w

π π π π π π→− →−

−= = =

− − − So, the limit can be done and we done with the problem! The limit is then,

( )24

sinlim

16 8w

ww

π π→−

= −−


2

ln 3limt

tt→∞

.

Step 1 The first step we should really do here is verify that L’Hospital’s Rule can in fact be used on this limit. This may seem like a silly step given that we are told to use L’Hospital’s Rule. However, in later sections we won’t be told to use it when/if it can be used. Therefore, we really need to get in the habit of checking that it can be used before applying it just to make sure that we can. If we apply L’Hospital’s Rule to a problem that it can’t be applied to then it’s is almost assured that we will get the wrong answer (it’s always possible you might get lucky and get the correct answer, but we will only be very lucky if it does). So a quick check shows us that,


Calculus I


( )2

ln 3as

tt

t∞

→ ∞ →∞


( ) 1

2 2

ln 3 1lim lim lim2 2

tt t t

tt t t→∞ →∞ →∞

= =

Don’t forget to simplify after taking the derivatives. This can often be the difference between being able to do the problem or not. Step 3 At this point all we need to do is try the limit and see if it can be done.

( )2 2

ln 3 1lim lim 02t t

tt t→∞ →∞

= = So, the limit can be done and we done with the problem! The limit is then,

( )2

ln 3lim 0t

tt→∞

=


( )

2

220

sin 2 7 2lim

1z

z z zz z→

+ −

+.



Calculus I


( )( )

2

22

sin 2 7 2 0as 001

z z zz

z z+ −

→ →+

and so this is a form that allows the use of L’Hospital’s Rule. Step 2 Before actually using L’Hospital’s Rule it might be better if we multiply out the denominator to make the derivative (and later steps a little easier). Doing this gives,

( )

( )( )2 2

2 4 3 220 0

sin 2 7 2 sin 2 7 2lim lim

21z z

z z z z z zz z zz z→ →

+ − + −=

+ ++

Now let’s apply L’Hospitals’s Rule.

( )( )

( )2

2 3 220 0

sin 2 7 2 2cos 2 14 2lim lim

4 6 21z z

z z z z zz z zz z→ →

+ − + −=

+ ++

Step 3 At this point let’s try the limit and see if it can be done. However, in this case, we can see that,

( )3 2

2cos 2 14 2 0as 04 6 2 0

z zz

z z z+ −

→ →+ +

Step 4 So, using L’Hospital’s Rule doesn’t give us a limit that we can do. However, the new limit is one that can use L’Hospital’s Rule on so let’s do that.

( )( )

( )2

2 220 0

sin 2 7 2 4sin 2 14 14lim lim12 12 2 21z z

z z z zz zz z→ →

+ − − += =

+ ++

Okay, the second L’Hospital’s Rule gives us a limit we can do and so the answer is,

( )( )

2

220

sin 2 7 2lim 7

1z

z z zz z→

+ −=

+

5. Use L’Hospital’s Rule to evaluate 2

1limx x

x−→−∞ e

.


Calculus I



2

1as xxx −

∞→ − ∞ →

∞e and so this is a form that allows the use of L’Hospital’s Rule. Step 2 So, at this point let’s just apply L’Hospital’s Rule.

2

1 1

2lim limx xx x

x x− −→−∞ →−∞

=−e e


1

2as xxx −

−∞→ − ∞ →

− −∞e Step 4 So, using L’Hospital’s Rule doesn’t give us a limit that we can do. However, the new limit is one that can use L’Hospital’s Rule on so let’s do that.

2

1 1 1

2 2lim lim lim 0x x xx x x

x x− − −→−∞ →−∞ →−∞

= = =−e e e

Okay, the second L’Hospital’s Rule gives us a limit we can do and so the answer is,

2

1lim 0x x

x−→−∞

=e

6. Use L’Hospital’s Rule to evaluate 2 4

lim2z

z

zz

z→∞

+−

ee

.


Calculus I



2 4

as2

z

zzz

z+ ∞

→ ∞ →− −∞

ee


2 4 42 4lim lim

2 2z z

z z

z zz z

z→∞ →∞

+ +=

− −e ee e


42 4as

2

z

zzz + ∞

→ ∞ →− −∞

ee

Step 4 So, using L’Hospital’s Rule doesn’t give us a limit that we can do. However, the new limit is one that can use L’Hospital’s Rule on so let’s do that.

2 4 4 42 4 2 16lim lim lim

2 2z z z

z z z

z z zz z

z→∞ →∞ →∞

+ + += =

− − −e e ee e e

Step 5 Now, at this point we need to be careful. It looks like we are still in a case of an infinity divided by an infinity and that looks to continue forever if we keep applying L’Hospital’s Rule. However, do not forget to do some basic simplifications where you can. If we simplify we get the following.


Calculus I


( ) ( ) ( )2 4

4 3lim lim 2 16 lim 2 162z z z

zz z z z

zz

z→∞ →∞ →∞

− −+= + − = − −

−e e e e ee

and this is something that we can take the limit of. So the answer is,

( )2 4

3lim lim 2 162z z

zz z

zz

z→∞ →∞

−+= − − = −∞

−e e ee

Again, it cannot be stressed enough that you’ve got to do simplification where you can. For some of these problems that can mean the difference between being able to do the problem or not.

7. Use L’Hospital’s Rule to evaluate 3lim ln 1

tt

t→∞

+ .

Step 1 The first thing to notice here is that is not in a form that allows L’Hospital’s Rule. L’Hospital’s Rule only works on a certain class of rational functions and this is clearly not a rational function. Note however that it is in the following indeterminate form,

( ) ( )3as ln 1 0t tt

→ ∞ + → ∞

and as we discussed in the notes for this section we can always turn this kind of indeterminate form into a rational expression that will allow L’Hospital’s Rule to be applied. Step 2 The real question is do we move the first term or the second term to the denominator. From the looks of things it appears that it would be best to move the first term to the denominator.

3ln 13lim ln 1 lim 1t t

ttt t

→∞ →∞

+ + =

Notice as well that,


Calculus I


3ln 10as 1 0

ttt

+ → ∞ →

and we can use L’Hospital’s Rule on this. Step 3 Applying L’Hospital’s Rule gives,

2

2

33ln 1 313lim ln 1 lim lim1 1t t t

tt tt

t t t→∞ →∞ →∞

− + + + = = −

Can you see why we chose to move the t to the denominator? Moving the logarithm would have left us with a very messy derivative to take! It might have ended up working okay for us, but the work would be greatly increased. Step 4 Do not forget to simplify after we’ve taken the derivative. This problem becomes very simple if we do that.

3ln 13 3lim ln 1 lim lim 31 31t t t

ttt t t

→∞ →∞ →∞

+ + = = = +

8. Use L’Hospital’s Rule to evaluate ( )2 2

0lim ln 4w

w w+→

.

Step 1 The first thing to notice here is that is not in a form that allows L’Hospital’s Rule. L’Hospital’s Rule only works on a certain class of rational functions and this is clearly not a rational function. Note however that it is in the following indeterminate form, ( ) ( ) ( )2 2as 0 ln 4 0w w w+→ → −∞ and as we discussed in the notes for this section we can always turn this kind of indeterminate form into a rational expression that will allow L’Hospital’s Rule to be applied. Step 2


Calculus I


The real question is do we move the first term or the second term to the denominator. From the looks of things it appears that it would be best to move the first term to the denominator.

( ) ( )22 2

0 02

ln 4lim ln 4 lim 1w w

ww w

w+ +→ →

=

Notice as well that,

( )2

2

ln 4as 0 1

ww

w

+ −∞→ →

∞


( ) ( )22 2

0 0 02 3

2ln 4lim ln 4 lim lim1 2w w w

w ww ww w

+ + +→ → → = = −

Can you see why we chose to move the first term to the denominator? Moving the logarithm would have left us with a very messy derivative to take! It might have ended up working okay for us, but the work would be greatly increased. Step 4 Do not forget to simplify after we’ve taken the derivative. This problem becomes very simple if we do that. In fact, it is the only way to actually get an answer for this problem. If we do not simplify will get stuck in a never ending chain of infinity divided by infinity forms no matter how many times we apply L’Hospital’s Rule.

( ) ( ) ( )2

2 2 2

0 0 02

ln 4lim ln 4 lim lim 01w w w

ww w w

w+ + +→ → →

= = − =

9. Use L’Hospital’s Rule to evaluate ( ) ( )21lim 1 tanx

x xπ+→

− .

Step 1 The first thing to notice here is that is not in a form that allows L’Hospital’s Rule. L’Hospital’s Rule only works on a certain class of rational functions and this is clearly not a rational function. Note however that it is in the following indeterminate form,


Calculus I


( ) ( ) ( ) ( )2as 1 1 tan 0x x xπ+→ − → ∞ and as we discussed in the notes for this section we can always turn this kind of indeterminate form into a rational expression that will allow L’Hospital’s Rule to be applied. Step 2 The real question is do we move the first term or the second term to the denominator. At first glance it might appear that neither term will be particularly useful in the denominator. In particular, if we move the tangent to the denominator we would end up needing to differentiate a

term in the form 1tan . That doesn’t look to be all that fun to differentiate and we’re liable to

end up with a mess when we are done. However, that is exactly the term we are going to move to the denominator for reasons that will quickly become apparent.

( ) ( )( ) ( )21 1 1

22

1 1lim 1 tan lim lim1 cottan

x x x

x xx xx

x

ππ

π+ + +→ → →

− − − = =

Step 3 With a little simplification after moving the tangent to the denominator we ended up with something that doesn’t look all that bad. We’ll also see that the remainder of this problem is going to be quite simple. Before we proceed however we should notice as well that,

( )2

1 0as 1cot 0

xxxπ

+ −→ →


( ) ( ) ( ) ( )2 21 1 12 2 2

1 1 2lim 1 tan lim limcot cscx x x

xx xx x

ππ π π π+ + +→ → →

− − = = = − −

10. Use L’Hospital’s Rule to evaluate ( ) 21

0lim cos 2y

yy+→

.

Step 1


Calculus I


The first thing to notice here is that is not in a form that allows L’Hospital’s Rule. L’Hospital’s Rule only works on certain classes of rational functions and this is clearly not a rational function. Note however that it is in the following indeterminate form,

( ) 21

as 0 cos 2 1yy y+ ∞→ → and as we discussed in the notes for this section we can do some manipulation on this to turn it into a problem that can be done with L’Hospital’s Rule. Step 2 First, let’s define,

( ) 21

cos 2 yz y= and take the log of both sides. We’ll also do a little simplification.

( )( ) ( ) ( )22

11

2

ln cos 2ln ln cos 2 ln cos 2y

y

yz y y

y = = =

Step 3 We can now take the limit as 0y +→ of this.

[ ] ( )20 0

ln cos 2lim ln limy y

yz

y+ +→ →

=

Before we proceed let’s notice that we have the following,

( ) ( )2

ln cos 2 ln 1 0as 00 0

yy

y+ → → =

and we have a limit that we can use L’Hospital’s Rule on. Step 4 Applying L’Hospital’s Rule gives,

[ ] ( )( )

( ) ( )20 0 0 0

2sin 2ln cos 2 cos 2 tan 2

lim ln lim lim lim2y y y y

yy y y

zy y y+ + + +→ → → →

− − = = =

Step 5 We now have a limit that behaves like,


Calculus I


( )tan 2 0as 00

yy

y+ −

→ →

and so we can use L’Hospital’s Rule on this as well. Doing this gives,

[ ] ( ) ( )2

0 0 0

tan 2 2sec 2lim ln lim lim 2

1y y y

y yz

y+ + +→ → →

− −= = = −

Step 6 Now all we need to do is recall that, ln zz = e This in turn means that we can do the original limit as follows,

( )[ ]

2 0ln1 2

0 0 0

limlnlim cos 2 lim lim y

zz

y y yyy z +→

+ + +

−

→ → →= = = = e e e

11. Use L’Hospital’s Rule to evaluate 1

limx

x xx→∞

+ e .

Step 1 The first thing to notice here is that is not in a form that allows L’Hospital’s Rule. L’Hospital’s Rule only works on certain classes of rational functions and this is clearly not a rational function. Note however that it is in the following indeterminate form,

1

0as x xx x → ∞ + → ∞ e and as we discussed in the notes for this section we can do some manipulation on this to turn it into a problem that can be done with L’Hospital’s Rule. Step 2 First, let’s define,

1x xz x = + e

and take the log of both sides. We’ll also do a little simplification.


Calculus I


1

1ln

ln ln lnx

x xxx

xz x x

x

+ = + = + =

ee e

Step 3 We can now take the limit as x → ∞ of this.

[ ]ln

lim ln limx x

x xz

x→∞ →∞

+ =

e

Before we proceed let’s notice that we have the following,

ln

asx x

xx

+ ∞ → ∞ →=∞

e

and we have a limit that we can use L’Hospital’s Rule on. Step 4 Applying L’Hospital’s Rule gives,

[ ]1ln 1lim ln lim lim lim1x x x x

xx xx

x

x xzx x→∞ →∞ →∞ →∞

+ + + + = = =+

ee eee

Step 5 We now have a limit that behaves like,

1asx

xxx

+ ∞→ ∞ →

+ ∞ee

and so we can use L’Hospital’s Rule on this as well. Doing this gives,

[ ] ( )1lim ln lim lim lim lim 1 11x x x x x

x x x

x x xzx→∞ →∞ →∞ →∞ →∞

+= = = = =

+ +e e ee e e

Notice that we did have to use L’Hospital’s Rule twice here and we also made sure to do some simplification so we could actually take the limit. Step 6 Now all we need to do is recall that, ln zz = e


Calculus I


This in turn means that we can do the original limit as follows,

[ ]1 lnlimlnlim lim lim x

zz

x x x

x xx z →∞

→∞ →∞ →∞ + = = = = e e e e

Linear Approximations 1. Find a linear approximation to ( ) 2 103 xf x x −= e at 5x = .

Step 1 We’ll need the derivative first as well as a couple of function evaluations. ( ) ( ) ( )2 10 2 103 6 5 15 5 33x xf x x f f− −′ ′= + = =e e Step 2 There really isn’t much to do at this point other than write down the linear approximation.

( ) ( )15 33 5 33 150L x x x= + − = − While it wasn’t asked for, here is a quick sketch of the function and the linear approximation.


Calculus I


2. Find a linear approximation to ( ) 4 36 3 7h t t t t= − + − at 3t = − .

Step 1 We’ll need the derivative first as well as a couple of function evaluations. ( ) ( ) ( )3 24 18 3 3 227 3 267h t t t h h′ ′= − + − = − = − Step 2 There really isn’t much to do at this point other than write down the linear approximation.

( ) ( )227 267 3 267 574L t t t= − + = − − While it wasn’t asked for, here is a quick sketch of the function and the linear approximation.

3. Find the linear approximation to ( ) 4g z z= at 2z = . Use the linear approximation to

approximate the value of 4 3 and 4 10 . Compare the approximated values to the exact values. Step 1 We’ll need the derivative first as well as a couple of function evaluations.

( ) ( ) ( ) ( )3 314 4 41 1

4 42 2 2 2g z z g g− −′ ′= = =

Step 2 Here is the linear approximation.


Calculus I


( ) ( )( )31

4 4142 2 2L z z−= + −

Step 3 Finally here are the approximations of the values along with the exact values.

( ) ( )( ) ( )3 1.33786 3 1.31607 % Error : 1.65523

10 2.37841 10 1.77828 % Error : 33.7481

L g

L g

= =

= =

So, as we might have expected the farther from 2z = we got the worse the approximation is. Recall that the approximation will generally be more accurate the closer to the point of the linear approximation. 4. Find the linear approximation to ( ) ( )cos 2f t t= at 1

2t = . Use the linear approximation to

approximate the value of ( )cos 1 and ( )cos 9 . Compare the approximated values to the exact

values. Step 1 We’ll need the derivative first as well as a couple of function evaluations. ( ) ( ) ( ) ( ) ( ) ( )1 1

2 22sin 2 cos 1 2sin 1f t t f f′ ′= − = = − Step 2 Here is the linear approximation.

( ) ( ) ( ) ( ) ( )1 12 2cos 1 2sin 1 0.5403 1.6829L t t t= − − = − −

Make sure your calculator is set in radians! Remember that we use radians by default in this class. Step 3 Finally here are the approximations of the values along with the exact values.

( ) ( )( ) ( )1 0.301169 1 0.416147 % Error : 27.6292

9 13.7647 9 0.660317 % Error : 2184.56

L f

L f

= − = −

= − =


Calculus I


So, as we might have expected the farther from 12t = we got the worse the approximation is.

Recall that the approximation will generally be more accurate the closer to the point of the linear approximation. 5. Without using any kind of computational aid use a linear approximation to estimate the value of 0.1e . Hint : This is really nothing more than Problem 3 and 4 from this section. The only difference is that you need to determine the function and the point for the linear approximation. The function should be pretty obvious given the value we are asked to estimate. There should also be a pretty obvious point to use given that we aren’t supposed to use calculators/computers. Step 1 This is really the same problem as Problems 3 & 4 from this section. The difference is that we need to determine the function and point for the linear approximation. Given the value we are being asked to estimate it should be fairly clear that the function should be,

( ) xf x = e

The point for the linear approximation should also be somewhat clear. With the function in hand it’s now clear that we are being asked to use a linear approximation to estimate ( )0.1f . So,

we’ll need a point that is close to 0.1x = and one that we can evaluate in the function without a calculator. It therefore seems fairly clear that 0x = would be a really nice point use for the linear approximation. Step 2 At this point finding the linear approximation shouldn’t be too bad so here is the work for that. ( ) ( ) ( )0 1 0 1xf x f f′ ′= = =e The linear approximation is then,

( ) ( ) ( )1 1 0 1L t x x= + − = + Step 3 The estimation of 0.1e is then,


Calculus I


( )0.1 0.1 1.1L≈ =e For comparison purposes the exact value is ( )0.1 1.10517f = and so we have an error of

0.467884 %.

Differentials 1. Compute the differential for ( ) ( )2 secf x x x= − .

Solution There is not really a whole lot to this problem.

( ) ( )( )2 sec tandf x x x dx= −

Don’t forget to tack on the dx at the end!

2. Compute the differential for 4 2 4x x xw − += e .


( ) 4 23 44 2 4 x x xdw x x dx− += − + e

Don’t forget to tack on the dx at the end! 3. Compute the differential for ( ) ( ) ( )ln 2 sin 2h z z z= .



Calculus I


( ) ( ) ( )1 sin 2 2ln 2 cos 2dh z z z dzz

= +

Don’t forget to tack on the dz at the end!

4. Compute dy and y∆ for 2xy = e as x changes from 3 to 3.01.

Step 1 First let’s get the actual change, y∆ .

2 23.01 3 501.927y∆ = − =e e

Step 2 Next, we’ll need the differential.

2

2 xdy x dx= e Step 3 As x changes from 3 to 3.01 we have 3.01 3 0.01x∆ = − = and we’ll assume that

0.01dx x≈ ∆ = . The approximate change, dy, is then,

( ) ( )232 3 0.01 486.185dy = =e Don’t forget to use the “starting” value of x (i.e. 3x = ) for all the x’s in the differential. 5. Compute dy and y∆ for 5 32 7y x x x= − + as x changes from 6 to 5.9. Step 1 First let’s get the actual change, y∆ .

( ) ( )( ) ( ) ( )( )5 3 5 35.9 2 5.9 7 5.9 6 2 6 7 6 606.215y∆ = − + − − + = − Step 2 Next, we’ll need the differential.


Calculus I


( )4 25 6 7dy x x dx= − + Step 3 As x changes from 6 to 5.9 we have 5.9 6 0.1x∆ = − = − and we’ll assume that 0.1dx x≈ ∆ = − . The approximate change, dy, is then,

( ) ( )( )( )4 25 6 6 6 7 0.1 627.1dy = − + − = −

Don’t forget to use the “starting” value of x (i.e. 6x = ) for all the x’s in the differential. 6. The sides of a cube are found to be 6 feet in length with a possible error of no more than 1.5 inches. What is the maximum possible error in the volume of the cube if we use this value of the length of the side to compute the volume? Step 1 Let’s get everything set up first. If we let the side of the cube be denoted by x the volume is then, ( ) 3V x x= We are told that 6x = and we can assume that 1.5

12 0.125dx x≈ ∆ = = (don’t forget to convert the inches to feet!). Step 2 We want to estimate the maximum error in the volume and so we can again assume that

V dV∆ ≈ . The differential is then, 23dV x dx= The maximum error in the volume is then,

( ) ( )2 33 6 0.125 13.5ftV dV∆ ≈ = =


Calculus I


Newton’s Method 1. Use Newton’s Method to determine 2x for ( ) 3 27 8 3f x x x x= − + − if 0 5x =

Step 1 There really isn’t that much to do with this problem. We know that the basic formula for Newton’s Method is,

( )( )1

nn n

n

f xx x

f x+ = −′

so all we need to do is run through this twice. Here is the derivative of the function since we’ll need that. ( ) 23 14 8f x x x′ = − + We just now need to run through the formula above twice. Step 2 The first iteration through the formula for 1x is,

( )( )

( )( )

01 0

0

5 135 5 65 13

f x fx x

ff x−

= − = − = − =′′

Step 3 The second iteration through the formula for 2x is,

( )( )

( )( )

12 1

1

6 96 6 5.718756 32

f x fx x

ff x= − = − = − =

′′

So, the answer for this problem is 2 5.71875x = .

Although it was not asked for in the problem statement the actual root is 5.68577952608963. Note as well that this did require some computational aid to get and it not something that you can, in general, get by hand.


Calculus I


2. Use Newton’s Method to determine 2x for ( ) ( ) 2cosf x x x x= − if 0 1x =

Step 1 There really isn’t that much to do with this problem. We know that the basic formula for Newton’s Method is,

( )( )1

nn n

n

f xx x

f x+ = −′

so all we need to do is run through this twice. Here is the derivative of the function since we’ll need that. ( ) ( ) ( )cos sin 2f x x x x x′ = − − We just now need to run through the formula above twice. Step 2 The first iteration through the formula for 1x is,

( )( )

( )( )

01 0

0

1 0.45969769411 1 0.80023294321 2.301168679

f x fx x

ff x−

= − = − = − =′ −′

Don’t forget that for us angles are always in radians so make sure your calculator is set to compute in radians. Step 3 The second iteration through the formula for 2x is,

( )( )

( )( )

12 1

1

0.80023294320.8002329432

0.8002329432

0.082978839480.8002329432 0.74409439851.478108132

f x fx x

ff x= − = −

′′

−= − =

−

So, the answer for this problem is 2 0.7440943985x = .


Calculus I


Although it was not asked for in the problem statement the actual root is 0.739085133215161. Note as well that this did require some computational aid to get and it not something that you can, in general, get by hand. 3. Use Newton’s Method to find the root of 4 35 9 3 0x x x− + + = accurate to six decimal places in the interval [ ]4,6 .

Step 1 First, recall that Newton’s Method solves equation in the form ( ) 0f x = and so it is (hopefully)

fairly clear that we have, ( ) 4 35 9 3f x x x x= − + + Next, we are not given a starting value, 0x , but we were given an interval in which the root exists

so we may as well use the midpoint of this interval as our starting point or, 0 5x = . Note that this

is not the only value we could use and if you use a different one (which is perfectly acceptable) then your values will be different from those here. At this point all we need to do is run through Newton’s Method,

( )( )1

nn n

n

f xx x

f x+ = −′

until the answers agree to six decimal places. Step 2 The first iteration through the formula for 1x is,

( )( )

01 0

0

485 4.641791045134

f xx x

f x= − = − =

′


( )( )

12 1

1

8.9505420574.641791045 4.53754395985.85891882

f xx x

f x= − = − =

′


Calculus I


We’ll need to keep going because even the first decimal is not correct yet. Step 4 The second iteration through the formula for 3x is,

( )( )

23 2

2

0.63299674134.537543959 4.52897372773.85993168

f xx x

f x= − = − =

′

At this point we are accurate to the first decimal place so we need to continue. Step 5 The second iteration through the formula for 4x is,

( )( )

34 3

3

0.0040661330054.528973727 4.5289179672.91199944

f xx x

f x= − = − =

′

At this point we are accurate to 4 decimal places so we need to continue. Step 6 The second iteration through the formula for 5x is,

( )( )

45 4

4

71.714694911*14.52891796 4.5289179672.905850

006

f xx x

f x

−

= − = − =′

At this point we are accurate to 8 decimal places which is actually better than we asked and so we can officially stop and we can estimate that the root in the interval is,

4.52891796x ≈ Using computational aids we found that the actual root in this interval is 4.52891795729. Note that this wasn’t actually asked for in the problem and is only given for comparison purposes. 4. Use Newton’s Method to find the root of 22 5 xx + = e accurate to six decimal places in the interval [ ]3, 4 .

Step 1


Calculus I


First, recall that Newton’s Method solves equation in the form ( ) 0f x = and so we’ll need move

everything to one side. Doing this gives, ( ) 22 5 xf x x= + −e Note that we could have just as easily gone the other direction. All that would have done was change the signs on the function and derivative evaluations in the work below. The final answers however would not be changed. Next, we are not given a starting value, 0x , but we were given an interval in which the root exists

so we may as well use the midpoint of this interval as our starting point or, 0 3.5x = . Note that

this is not the only value we could use and if you use a different one (which is perfectly acceptable) then your values will be different that those here. At this point all we need to do is run through Newton’s Method,

( )( )1

nn n

n

f xx x

f x+ = −′

until the answers agree to six decimal places. Step 2 The first iteration through the formula for 1x is,

( )( )

01 0

0

3.6154519593.5 3.31086233419.11545196

f xx x

f x−

= − = − =−′


( )( )

12 1

1

0.48513199923.310862334 3.27661442214.16530146

f xx x

f x−

= − = − =−′

We’ll need to keep going because even the first decimal is not correct yet. Step 4 The second iteration through the formula for 3x is,


Calculus I


( )( )

23 2

2

0.01354634863.276614422 3.27560195113.37949281

f xx x

f x−

= − = − =−′

At this point we are accurate to two decimal places so we need to continue. Step 5 The second iteration through the formula for 4x is,

( )( )

34 3

3

0.000011520565963.27513.3

601951 3.27560108956740003

f xx x

f x−

= − = − =−′

At this point we are accurate to 6 decimal places which is what we were asked to do and so we can officially stop and we can estimate that the root in the interval is,

3.275601089x ≈ Using computational aids we found that the actual root in this interval is 3.27560108884732 . Note that this wasn’t actually asked for in the problem and is only given for comparison purposes and it does look like Newton’s Method did a pretty good job as this is identical to the final iteration that we did. 5. Use Newton’s Method to find all the roots of 3 2 15 1 0x x x− − + = accurate to six decimal places. Hint : Can you use your knowledge of Algebra to determine how many roots this equation should have? Maybe a graph of the function could also be useful for this problem. Step 1 First, recall that Newton’s Method solves equation in the form ( ) 0f x = and so it is (hopefully)

fairly clear that we have, ( ) 3 2 15 1f x x x x= − − + Next, we are not given a starting value, 0x and unlike Problems 3 & 4 above we are not even

given an interval to use as a way to determine a good possible value of 0x . We are also not even

told how many roots we need to find.


Calculus I


Of course, if we recall our Algebra skills we can see that we have a cubic polynomial and so there should be at most three distinct roots of the equation (there may be some that repeat and so we may not have three distinct roots…). Knowing this all we really need to do to get potential starting values is to do a quick sketch of the function. In determining a proper range of x values just keep in mind what we know about limits at infinity. Because the largest power of x is odd in this case we know that as x → ∞ the graph should also be approaching positive infinity and as x → −∞ the graph should be approaching negative infinity. So, we can start with a large range of x’s that gives the behavior we expect at the right/left ends of the graph and then narrow it down until we see the actual roots showing up on the graph. Doing this gives,

So, it looks like we are going to have three roots here (i.e. the graph crosses the x-axis three times and so three roots…).

For each root we’ll use the graph to pick a value of 0x that is close to the root we are after (we’ll

go from left to right for the problem) and then run through Newton’s Method,

( )( )1

nn n

n

f xx x

f x+ = −′

until the answers agree to six decimal places. Note as well that unlike Problems 3 & 4 we are not going to put in all the function evaluations for this problem. We’ll leave that to you to check and verify our final answers for each iteration.


Calculus I


Step 2 For the left most root let’s start with 0 3.5x = − . Here are the results of iterating through

Newton’s Method for this root.

( )( )( )( )( )( )

01 0

0

12 1

1

23 2

2

3.443478261 No decimal places agree

3.442146902 Accurate to two decimal places

3.44214617 Accurate to six decimal places

f xx x

f x

f xx x

f x

f xx x

f x

= − = −′

= − = −′

= − = −′

So, it looks like the estimate of the left most root is : 3.44214617x ≈ − . Step 3 For the middle root let’s start with 0 0x = . Be careful with this root. From the graph we may be

tempted to just say the root is zero. However, as we’ll see the root is not zero. It is close to zero, but is not exactly zero! Here are the results of iterating through Newton’s Method for this root.

( )( )( )( )( )( )

01 0

0

12 1

1

23 2

2


0.06639231824 Accurate to three decimal places

0.06639231426 Accurate to eight decimal places

f xx x

f x

f xx x

f x

f xx x

f x

= − =′

= − =′

= − =′

So, it looks like the estimate of the middle root is : 0.06639231426x ≈ . Step 4 For the right most root let’s start with 0 4.5x = . Here are the results of iterating through



Calculus I


( )( )( )( )( )( )( )( )

01 0

0

12 1

1

23 2

2

34 3

3


4.375763556 Accurate to one decimal place

4.375753856 Accurate to four decimal places

4.375753856 Accurate to nine

f xx x

f x

f xx x

f x

f xx x

f x

f xx x

f x

= − =′

= − =′

= − =′

= − =′

decimal places

So, it looks like the estimate of the right most root is : 4.375753856x ≈ . Step 5 Using computational aids we found that the actual roots of this equation to be, 3.44214616993 0.0663923142603 4.37575385567x x x= − = = Note that these weren’t actually asked for in the problem and are only given for comparison purposes. As a final warning about Newton’s Method, be careful to not assume that you’ll get six (or better in some cases) decimal places of accuracy with just a few iterations. These problems were chosen with the understanding that it would only take a few iterations of the method. There are problems and/or choices of 0x for which it will take significantly more

iterations to get any kind of real accuracy, provided the method even works for that equation and/or choice of 0x . Recall that we saw an example in the notes in which the method failed

spectacularly. 6. Use Newton’s Method to find all the roots of ( )22 sinx x− = accurate to six decimal places.

Hint : Can you use your knowledge what the graph of the left side and right side of this equation to determine how many roots this equation should have? Maybe a graph of the functions on the left and right side could also be useful for this problem. Step 1


Calculus I


First, recall that Newton’s Method solves equation in the form ( ) 0f x = and so we’ll need move

everything to one side. Doing this gives, ( ) ( )22 sinf x x x= − − Note that we could have just as easily gone the other direction. All that would have done was change the signs on the function and derivative evaluations in the work below. The final answers however would not be changed. Next, we are not given a starting value, 0x and unlike Problems 3 & 4 above we not even given

an interval to use as a way to determine a good possible value of 0x . We are also not even told

how many roots we need to find. So to estimate the number of roots of the equation let’s take a look at each side of the equation and realize that each root will in fact be the point of intersection of the two curves on the left and right of the equal sign. The left side of the original equation is a quadratic that will have its vertex at 2x = and open downward while the right side is the sine function. Given what we know of these two functions we should expect there to be at most two roots where the quadratic, on its way down, intersects with the sine function. Because the quadratic will never turn around and start moving back upwards it should never intersect with the sine function again after those points. So, let’s graph both the quadratic and sine function to see if our intuition on this is correct. Doing this gives,

So, it looks like we guessed correctly and should have two roots here.


Calculus I


For each root we’ll use the graph to pick a value of 0x that is close to the root we are after (we’ll

go from left to right for the problem) and then run through Newton’s Method,

( )( )1

nn n

n

f xx x

f x+ = −′

until the answers agree to six decimal places. Note as well that unlike Problems 3 & 4 we are not going to put in all the function evaluations for this problem. We’ll leave that to you to check and verify our final answers for each iteration. Also note that the analysis that we had to do to estimate the number of roots is something that does need to be done for these kinds of problems and it will differ for each equation. However, if you do have a basic knowledge of how most of the basic functions behave you can do this for most equations you’ll be asked to deal with. Step 2 For the left most root let’s start with 0 1.5x = − . Here are the results of iterating through


( )( )( )( )( )( )( )( )

01 0

0

12 1

1

23 2

2

34 3

3


1.728754674 Accurate to one decimal place

1.728466353 Accurate to three decimal places

1.728466319 Accurate to

f xx x

f x

f xx x

f x

f xx x

f x

f xx x

f x

= − = −′

= − = −′

= − = −′

= − = −′

seven decimal places

So, it looks like the estimate of the left most root is : 1.728466319x ≈ − . Step 3 For the right most root let’s start with 0 1x = . Here are the results of iterating through Newton’s

Method for this root.


Calculus I


( )( )( )( )( )( )

01 0

0

12 1

1

23 2

2


1.061549933 Accurate to two decimal places

1.061549775 Accurate to six decimal places

f xx x

f x

f xx x

f x

f xx x

f x

= − =′

= − =′

= − =′

So, it looks like the estimate of the right most root is : 1.061549775x ≈ . Step 5 Using computational aids we found that the actual roots of this equation to be, 1.72846631899718 1.06154977463138x x= − = Note that these weren’t actually asked for in the problem and are only given for comparison purposes. As a final warning about Newton’s Method, be careful to not assume that you’ll get six (or better in some cases) decimal places of accuracy with just a few iterations. These problems were chosen with the understanding that it would only take a few iterations of the method. There are problems and/or choices of 0x for which it will take significantly more

iterations to get any kind of real accuracy, provided the method even works for that equation and/or choice of 0x . Recall that we saw an example in the notes in which the method failed

spectacularly.


330,000,000 360,000 750P x x x x= − + − How many widgets should they try to sell in order to maximize their profit?


Calculus I


Step 1 Because these are essentially the same type of problems that we did in the Absolute Extrema section we will not be doing a lot of explanation to the steps here. If you need some practice on absolute extrema problems you should check out some of the examples and/or practice problems there. All we really need to do here is determine the absolute maximum of the profit function and the value of x that will give the absolute maximum. Here is the derivative of the profit function and the critical point(s) since we’ll need those for this problem. ( ) ( ) ( )2360,000 1500 1200 300 0 300, 1200P x x x x x x x′ = − + − = − − − = ⇒ = = Step 2 From the problem statement we can see that we only want critical points that are in the interval

[ ]0,1500 . As we can see both of the critical points from the above step are in this interval and so

we’ll need both of them. Step 3 The next step is to evaluate the profit function at the critical points from the second step and at the end points of the given interval. Here are those function evaluations.

( ) ( )( ) ( )0 30,000,000 300 19,500,000

1200 102,000,000 1500 52,500,000

P P

P P

= = −

= =

Step 4 From these evaluations we can see that they will need to sell 1200 widgets to maximize the profits. 2. A management company is going to build a new apartment complex. They know that if the complex contains x apartments the maintenance costs for the building, landscaping etc. will be, ( ) 24000 14 0.04C x x x= + − The land they have purchased can hold a complex of at most 500 apartments. How many apartments should the complex have in order to minimize the maintenance costs? Step 1


Calculus I


Because these are essentially the same type of problems that we did in the Absolute Extrema section we will not be doing a lot of explanation to the steps here. If you need some practice on absolute extrema problems you should check out some of the examples and/or practice problems there. All we really need to do here is determine the absolute minimum of the maintenance function and the value of x that will give the absolute minimum. Here is the derivative of the maintenance function and the critical point(s) since we’ll need those for this problem. ( ) 14 0.08 175C x x x′ = − = ⇒ = Step 2 From the problem statement we can see that we only want critical points that are in the interval

[ ]0,500 . As we can see both of the critical points from the above step are in this interval and so

we’ll need both of them. Step 3 The next step is to evaluate the maintenance function at the critical points from the second step and at the end points of the given interval. Here are those function evaluations. ( ) ( ) ( )0 4000 175 5225 500 1000C C C= = = Step 4 From these evaluations we can see that the complex should have 500 apartments to minimize the maintenance costs. 3. The production costs, in dollars, per day of producing x widgets is given by, ( ) 2 31750 6 0.04 0.0003C x x x x= + − + What is the marginal cost when 175x = and 300x = ? What do your answers tell you about the production costs? Step 1 From the notes in this section we know that the marginal cost is simply the derivative of the cost function so let’s start with that. ( ) 26 0.08 0.0009C x x x′ = − +


Calculus I


Step 2 The marginal costs for each value of x is then,

( ) ( )125 10.0625 300 63C C′ ′= = Step 3 From these computations we can see that is will cost approximately $10.06 to produce the 126th widget and approximately $63 to produce the 301st widget. 4. The production costs, in dollars, per month of producing x widgets is given by,

( ) 10000200 0.5C x xx

= + +

What is the marginal cost when 200x = and 500x = ? What do your answers tell you about the production costs? Step 1 From the notes in this section we know that the marginal cost is simply the derivative of the cost function so let’s start with that.

( ) 2

100000.5C xx

′ = −

Step 2 The marginal costs for each value of x is then,

( ) ( )200 0.25 500 0.46C C′ ′= = Step 3 From these computations we can see that is will cost approximately 25 cents to produce the 201st widget and approximately 46 cents to produce the 501st widget. 5. The production costs, in dollars, per week of producing x widgets is given by, ( ) 2 34000 32 0.08 0.00006C x x x x= − + +


Calculus I


and the demand function for the widgets is given by, ( ) 2250 0.02 0.001p x x x= + − What is the marginal cost, marginal revenue and marginal profit when 200x = and 400x = ? What do these numbers tell you about the cost, revenue and profit? Step 1 First we need to get the revenue and profit functions. From the notes for this section we know that these functions are,

( ) ( )

( ) ( ) ( )

2 3

2 3

Revenue : 250 0.02 0.001

Profit : 4000 282 0.06 0.00106

R x x p x x x x

P x R x C x x x x

= = + −

= − = − + − −

Step 2 From the notes in this section we know that the marginal cost, marginal revenue and marginal profit functions are simply the derivative of the cost, revenue and profit functions so let’s start with those.

( )( )( )

2

2

2

32 0.16 0.00018

250 0.04 0.003

282 0.12 0.00318

C x x x

R x x x

P x x x

′ = − + +

′ = + −

′ = − −

Step 3 The marginal cost, marginal revenue and marginal profit for each value of x is then,

( ) ( ) ( )( ) ( ) ( )200 7.2 200 138 200 130.8

400 60.8 400 214 400 274.8

C R P

C R P

′ ′ ′= = =

′ ′ ′= = − = −

Step 4 From these computations we can see that producing the 201st widget will cost approximately $7.2 and will add approximately $138 in revenue and $130.8 in profit. Likewise, producing the 401st widget will cost approximately $60.8 and will see a decrease of approximately $214 in revenue and a decrease of $274.8 in profit.


Calculus I



CALCULUS I Assignment Problems


Paul Dawkins

Calculus I



Introduction .............................................................................................................................................. 2 Rates of Change ......................................................................................................................................... 3 Critical Points ............................................................................................................................................ 3 Minimum and Maximum Values .............................................................................................................. 7 Finding Absolute Extrema...................................................................................................................... 10 The Shape of a Graph, Part I ................................................................................................................... 12 The Shape of a Graph, Part II ................................................................................................................. 19 The Mean Value Theorem ...................................................................................................................... 24 Optimization ........................................................................................................................................... 25 More Optimization Problems................................................................................................................. 27 Indeterminate Forms and L’Hospital’s Rule ......................................................................................... 29 Linear Approximations .......................................................................................................................... 31 Differentials............................................................................................................................................. 32 Newton’s Method .................................................................................................................................... 33 Business Applications ............................................................................................................................ 34


Calculus I


Preface Here are a set of problems for my Calculus I notes. These problems do not have any solutions available on this site. These are intended mostly for instructors who might want a set of problems to assign for turning in. I try to put up both practice problems (with solutions available) and these problems at the same time so that both will be available to anyone who wishes to use them.


Introduction Here are a set of problems for which no solutions are available. The main intent of these problems is to have a set of problems available for any instructors who are looking for some extra problems. Note that some sections will have more problems than others and some will have more or less of a variety of problems. Most sections should have a range of difficulty levels in the problems although this will vary from section to section. Here is a list of topics in this chapter that have problems written for them. Rates of Change Critical Points Minimum and Maximum Values Finding Absolute Extrema The Shape of a Graph, Part I The Shape of a Graph, Part II The Mean Value Theorem Optimization Problems More Optimization Problems L’Hospital’s Rule and Indeterminate Forms Linear Approximations Differentials Newton’s Method Business Applications


Calculus I


Rates of Change As noted in the text for this section the purpose of this section is only to remind you of certain types of applications that were discussed in the previous chapter. As such there aren’t any problems written for this section. Instead here is a list of links (note that these will only be active links in the web version and not the pdf version) to problems from the relevant sections from the previous chapter. Each of the following sections has a selection of increasing/decreasing problems towards the bottom of the problem set.



Critical Points Determine the critical points of each of the following functions. Note that a couple of the problems involve equations that may not be easily solved by hand and as such may require some computational aids. These are marked are noted below. 1. ( ) 3 28 18 240 2R x x x x= − − +

2. ( ) 4 3 22 16 20 7f z z z z= − + −

3. ( ) 5 6 790

78 12 25g z z z z= − − +

4. ( ) 4 3 23 20 132 672 4g t t t t t= − − + − Note : Depending upon your factoring skills this may require some computational aids. 5. ( ) 2 3 4 515

210 15h x x x x x= − + − Note : Depending upon your factoring skills this may require some computational aids. 6. ( ) 3 24 7 1P w w w w= − − −

7. ( ) 3 27 3 15A t t t t= − + −


Calculus I


8. ( ) 2 3 44 2 6 3a t t t t= − − −

9. ( ) 4 3 23 20 6 120 5f x x x x x= − + + + Note : This problem will require some computational aids. 10. ( ) 5 4 310 15h v v v v= + + −

11. ( ) ( ) ( )5 43 2 1g z z z= − +

12. ( ) ( ) ( )24 22 8R q q q= + −

13. ( ) ( ) ( )23 22 1f t t t= − +

14. ( )2 2 13 5

w wf ww+ +

=−

15. ( ) 2

3 41th t

t−

=+

16. ( )2

2 3 8y yR y

y y−

=+ +

17. ( ) 3 7Y x x= −

18. ( ) ( )2

3 325f t t t= −

19. ( ) ( )25 2 8h x x x= +

20. ( ) ( ) 32 26 4Q w w w= − −

21. ( ) ( )47sin 2tQ t = −

22. ( ) ( )3cos 2 5g x x x= −


Calculus I


23. ( ) ( )7cos 2f x x x= +

24. ( ) ( )6sin 2 12h t t t= +

25. ( ) ( )3

5cos zw z =

26. ( ) ( )tan 4U z z z= −

27. ( ) ( ) ( )cos sinh x x x x= −

28. ( ) ( ) ( )2cos cos 2h x x x= −

29. ( ) ( ) ( )2 4cos cosf w w w= −

30. ( ) 14 3wF w += e

31. ( ) 2 1 zg z z −= e

32. ( ) ( ) 2

3 2 xA x x= − e

33. ( ) ( ) 286 1 t tP t t −= + e

34. ( ) 2 23 2 4x xf x + −= −e e

35. ( ) 2 24 8 2z z z zf z − −= +e e

36. ( ) 3 26 8y yh y −= e

37. ( ) 3 22 4t t tg t + −= e

38. ( ) ( )2ln 3Z t t t= + +

39. ( ) ( )2ln 1G r r r= − +


Calculus I


40. ( ) ( )2 6 ln 8 1A z z z= − + +

41. ( ) ( )24ln 2f x x x x= − + +

42. ( ) ( ) ( )ln 4 2 ln 4g x x x= + − +

43. ( ) ( ) ( )2ln 1 ln 4h t t t t= − + + −

44. The graph of some function, ( )f x , is shown. Based on the graph, estimate the location of

all the critical points of the function.






Calculus I







Calculus I


4. Sketch the graph of ( ) 123f x x= − and identify all the relative extrema and absolute extrema


(b) [ ]3,2−

(c) [ )4,1−

(d) ( )0,5

5. Sketch the graph of ( ) ( )22 1g x x= − + and identify all the relative extrema and absolute


(b) [ ]0,3

(c) [ ]1,5−

(d) [ ]1,1−

(e) [ )1,3

(f) ( )2, 4

6. Sketch the graph of ( ) 3 xh x −= e and identify all the relative extrema and absolute extrema of

the function on each of the following intervals. (a) ( ),−∞ ∞

(b) [ ]1,3−

(c) [ ]6, 1− −

(d) ( ]1,4

7. Sketch the graph of ( ) ( )cos 2h x x= + and identify all the relative extrema and absolute

extrema of the function on each of the following intervals. Do, all work for this problem in radians.

(a) ( ),−∞ ∞

(b) ,3 4π π −

(c) , 22π

π −

(d) 1 ,12


Calculus I


8. Sketch the graph of a function on the interval [ ]3,9 that has an absolute maximum at 5x =

and an absolute minimum at 4x = . 9. Sketch the graph of a function on the interval [ ]0,10 that has an absolute minimum at 5x =

and an absolute maximums at 0x = and 10x = . 10. Sketch the graph of a function on the interval ( ),−∞ ∞ that has a relative minimum at

7x = − , a relative maximum at 2x = and no absolute extrema. 11. Sketch the graph of a function that meets the following conditions :

(a) Has at least one absolute maximum. (b) Has one relative minimum. (c) Has no absolute minimum.

12. Sketch the graph of a function that meets the following conditions :

(a) Graphed on the interval [ ]2,9 .

(b) Has a discontinuity at some point interior to the interval. (c) Has an absolute maximum at the discontinuity in part (b). (d) Has an absolute minimum at the discontinuity in part (b).


(a) Graphed on the interval [ ]4,10− .

(b) Has no relative extrema. (c) Has an absolute maximum at one end point. (d) Has an absolute minimum at the other end point.


(a) Has a discontinuity at some point. (b) Has an absolute maximum and an absolute minimum. (c) Neither absolute extrema occurs at the discontinuity.

Finding Absolute Extrema For each of the following problems determine the absolute extrema of the given function on the specified interval. 1. ( ) 4 3 22 16 20 7f z z z z= − + − on [ ]2,6−


Calculus I


2. ( ) 4 3 22 16 20 7f z z z z= − + − on [ ]2,4−

3. ( ) 4 3 22 16 20 7f z z z z= − + − on [ ]0,2

4. ( ) 3 4 520 280 75 12Q w w w w= + + − on [ ]3,2−

5. ( ) 3 4 520 280 75 12Q w w w w= + + − on [ ]1,8−

6. ( ) 5 6 790

78 12 25g z z z z= − − + on [ ]1,1−

7. ( ) 4 3 23 20 132 672 4g t t t t t= − − + − on [ ]5,8− Note : Depending upon your factoring skills this may require some computational aids. 8. ( ) 4 3 23 20 132 672 4g t t t t t= − − + − on [ ]2,8− Note : Depending upon your factoring skills this may require some computational aids. 9. ( ) 3 214 11 4 3V x x x x= + − + on [ ]1,1−

10. ( ) 2 3 44 2 6 3a t t t t= − − − on [ ]2,1−

11. ( ) 2 38 3 7h x x x x= + + − on [ ]1,5−

12. ( ) 4 3 23 20 6 120 5f x x x x x= − + + + on [ ]1,5− Note : This problem will require some computational aids. 13. ( ) 5 4 310 15h v v v v= + + − on [ ]3,2−

14. ( ) ( ) ( )5 43 2 1g z z z= − + on [ ]1,3−

15. ( ) ( ) ( )24 22 8R q q q= + − on [ ]4,1−

16. ( ) 2

3 41th t

t−

=+

on [ ]2,4−

17. ( )2

2

6 91

x xg xx x

+ +=

+ + on [ ]6,0−


Calculus I


18. ( ) ( )2

3 325f t t t= − on [ ]2,6

19. ( ) ( )2

252 1F x t t t= + + + on [ ]2,1−

20. ( ) ( ) 32 26 4Q w w w= − − on [ ]125,−

21. ( ) ( )3cos 2 5g x x x= − on [ ]0,6

22. ( ) ( )33 10sin ws w w= − on [ ]10,38

23. ( ) ( )7cos 2f x x x= + on [ ]5,4−

24. ( ) ( ) ( )cos sinh x x x x= − on [ ]15, 5− −

25. ( ) 2 1 zg z z −= e on 51

2 2,−

26. ( ) ( ) 286 1 t tP t t −= + e on [ ]1,3−

27. ( ) 5 9 1 3 6x xf x + −= + +e e on [ ]1,0−

28. ( ) 3 26 8y yh y −= e on [ ]12 ,1−

29. ( ) ( )2ln 3Z t t t= + + on [ ]2,2−

30. ( ) ( )24ln 2f x x x x= − + + on [ ]1,9−

31. ( ) ( ) ( )2ln 1 ln 4h t t t t= − + + − on [ ]1,3

The Shape of a Graph, Part I For problems 1 – 4 the graph of a function is given. Determine the open intervals on which the function increases and decreases.


Calculus I


1.

2.

3.

4.


Calculus I


For problems 5 – 7 the graph of the derivative of a function is given. From this graph determine the open intervals in which the function increases and decreases. 5.

6.


Calculus I


7.

For problems 8 – 10 The known information about the derivative of a function is given. From this information answer each of the following questions.


8.

( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )1 0 3 0 8 0

0 on ,1 , 3,8 0 on 1,3 , 8,

f f f

f x f x

′ ′ ′= = =

′ ′< −∞ > ∞

9.

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )2 0 0 0 3 0 6 0

0 on 0,3 , 6, 0 on , 2 , 2,0 , 3,6

g g g g

g x g x

′ ′ ′ ′− = = = =

′ ′< ∞ > −∞ − −

10.

( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )1 0 2 0 5 0

0 on , 1 , 1,2 0 on 2,5 , 5,

h h h

h x h x

′ ′ ′− = = =

′ ′< −∞ − − > ∞



11. ( ) 3 215 63 3f t t t t= − + +

12. ( ) 2 3 420 8 4g x x x x= + + −


Calculus I


13. ( ) 3 28 18 24 10Q w w w w= − − −

14. ( ) 5 4 35

4 20 7f x x x x= + − −

15. ( ) 2 35 4 9 3P x x x x= − − −

16. ( ) 5 4 36 5R z z z z= + − +

17. ( ) 2 3 41 12 9 2h z z z z= − − −

18. ( ) ( )7 sin 4Q t t t= − + on 3 3

2 2,−

19. ( ) ( )26 20cos zf z z= − on [ ]0,22

20. ( ) ( )324cos 8 2xg x x= + + on [ ]30,25−

21. ( ) ( )9 5sin 2h w w w= − on [ ]5,0−

22. ( ) ( )5 7h x x x= +

23. ( ) ( )( )2

2 310 2W z w w= − +

24. ( ) ( ) 32 28 4f t t t= − −

25. ( )3 21

3 3x x xf x − −= e

26. ( ) ( )2 38 zh z z −= − e

27. ( ) ( )2ln 5 8A t t t= + +

28. ( ) ( )23 ln 1g x x x x= − + + +

29. Answer each of the following questions.

(a) What is the minimum degree of a polynomial that has exactly one relative extrema?


Calculus I


(b) What is the minimum degree of a polynomial that has exactly two relative extrema? (c) What is the minimum degree of a polynomial that has exactly three relative extrema? (d) What is the minimum degree of a polynomial that has exactly n relative extrema?

30. For some function, ( )f x , it is known that there is a relative minimum at 4x = − . Answer

each of the following questions about this function. (a) What is the simplest form that the derivative of this function? Note : There really are many

possible forms of the derivative so to make the rest of this problem as simple as possible you will want to use the simplest form of the derivative.

(b) Using your answer from (a) determine the most general form that the function itself can take.

(c) Given that ( )4 6f − = find a function that will have a relative minimum at 4x = − . Note

: There are many possible answers here so just give one of them. 31. For some function, ( )f x , it is known that there is a relative maximum at 1x = − . Answer

each of the following questions about this function. (a) What is the simplest form that the derivative of this function? Note : There really are many



(c) Given that ( )1 3f − = find a function that will have a relative maximum at 1x = − . Note

: There are many possible answers here so just give one of them. 32. For some function, ( )f x , it is known that there is a critical point at 3x = that is neither a

relative minimum or a relative maximum. Answer each of the following questions about this function.

(a) What is the simplest form that the derivative of this function? Note : There really are many possible forms of the derivative so to make the rest of this problem as simple as possible you will want to use the simplest form of the derivative.


(c) Given that ( )3 2f = find a function that will have a critical point at 3x = that is neither

a relative minimum or a relative maximum. Note : There are many possible answers here so just give one of them.

33. For some function, ( )f x , it is known that there is a relative maximum at 1x = and a

relative minimum at 4x = . Answer each of the following questions about this function. (a) What is the simplest form that the derivative of this function? Note : There really are many



Calculus I



(c) Given that ( )1 6f = and ( )4 2f = − find a function that will have a relative maximum

at 1x = and a relative minimum at 4x = . Note : There are many possible answers here so just give one of them.

34. Given that ( )f x and ( )g x are increasing functions will ( ) ( ) ( )h x f x g x= − always be

an increasing function? If so, prove that ( )h x will be an increasing function. If not, find

increasing functions, ( )f x and ( )g x , so that ( )h x will be a decreasing function and find a

different set of increasing functions so that ( )h x will be an increasing function.

35. Given that ( )f x is an increasing function. There are several possible conditions that we can

impose on ( )g x so that ( ) ( ) ( )h x f x g x= − will be an increasing function. Determine as

many of these possible conditions as you can. 36. For a function ( )f x determine a set of conditions on ( )f x , different from those given in

#15 in the practice problems, for which ( ) ( ) 2h x f x= will be an increasing function.

37. For a function ( )f x determine a single condition on ( )f x for which ( ) ( ) 3h x f x= will

be an increasing function. 38. Given that ( )f x and ( )g x are positive functions. Determine a set of conditions on them

for which ( ) ( ) ( )h x f x g x= will be an increasing function. Note that there are several

possible sets of conditions here, but try to determine the “simplest” set of conditions.

39. Repeat #38 for ( ) ( )( )

f xh x

g x= .

40. Given that ( )f x and ( )g x are increasing functions prove that ( ) ( )( )h x f g x= will also

be an increasing function.


Calculus I


The Shape of a Graph, Part II For problems 1 & 2 the graph of a function is given. Determine the open intervals on which the function is concave up and concave down. 1.

2.

For problems 3 – 5 the graph of the 2nd derivative of a function is given. From this graph determine the open intervals in which the function is concave up and concave down. 3.


Calculus I


4.

5.


Calculus I



(a) Determine the open intervals on which the function is concave up and concave down. (b) Determine the inflection points of the function.

6. ( ) 3 29 24 6f x x x x= + + −

7. ( ) 4 3 22 120 84 35Q t t t t t= − − − +

8. ( ) 5 4 33 20 40h z z z z= − +

9. ( ) 4 3 25 2 18 108 12g w w w w w= − − + −

10. ( ) 4 5 610 360 20 3g x x x x x= + + + −

11. ( ) ( )2

49 3 160sin xA x x x= − − on [ ]20,10−

12. ( ) ( ) 23cos 2 14f x x x= − − on [ ]0,6

13. ( ) ( )21 2 sin 2h t t t= + − on [ ]2,4−

14. ( ) ( )138R v v v= −

15. ( ) ( )( )251 3g x x x= − +

16. ( ) 4x xf x −= −e e

17. ( ) 2 wh w w −= e

18. ( ) ( )2 2ln 1A w w w= − +


(a) Identify the critical points of the function. (b) Determine the open intervals on which the function increases and decreases. (c) Classify the critical points as relative maximums, relative minimums or neither. (d) Determine the open intervals on which the function is concave up and concave down. (e) Determine the inflection points of the function.


Calculus I


(f) Use the information from steps (a) – (e) to sketch the graph of the function.

19. ( ) 2 310 30 2f x x x= − +

20. ( ) 3 314 4G t t t= + −

21. ( ) 4 3 24 18 9h w w w w= + − −

22. ( ) 3 4 510 10 3g z z z z= + +

23. ( ) 6 5 49 20 10f z z z z= − + +

24. ( ) ( )3 5sin 2Q t t t= − on [ ]1, 4−

25. ( ) ( )1 1

2 3cosg x x x= + on [ ]25,0−

26. ( ) ( )134h x x x= −

27. ( ) 2 1f t t t= +

28. ( ) ( )45 27A z z z= −

29. ( ) 4 6w wg w = −e e

30. ( )21

413 tP t t −= e

31. ( ) ( )31 xg x x −= + e

32. ( ) ( )2ln 1h z z z= + +

33. ( ) ( )22 8ln 4f w w w= − +

34. Answer each of the following questions.

(a) What is the minimum degree of a polynomial that has exactly two inflection points. (b) What is the minimum degree of a polynomial that has exactly three inflection points.


Calculus I


(c) What is the minimum degree of a polynomial that has exactly n inflection points. 35. For some function, ( )f x , it is known that there is an inflection point at 3x = . Answer each

of the following questions about this function. (a) What is the simplest form that the 2nd derivative of this function? . (b) Using your answer from (a) determine the most general form that the function itself can

take. (c) Given that ( )0 6f = − and ( )3 1f = find a function that will have an inflection point at

3x = . For problems 36 – 39 ( )f x is a polynomial. Given the 2nd derivative of the function classify, if

possible, each of the given critical points as relative minimums or relative maximum. If it is not possible to classify the critical point(s) clearly explain why they cannot be classified. 36. ( ) 23 4 15f x x x′′ = − − . The critical points are : 3x = − , 0x = and 5x = .

37. ( ) 3 24 21 24 68f x x x x′′ = − − + . The critical points are : 2x = − , 4x = and 7x = .

38. ( ) 2 323 18 9 4f x x x x′′ = + − − . The critical points are : 4x = − , 1x = − and 3x = .

39. ( ) 2 3 4216 410 249 60 5f x x x x x′′ = − + − + . The critical points are : 1x = , 4x = and

5x = .

40. Use ( ) ( ) ( )3 41 1f x x x= + − for this problem.

(a) Determine the critical points for the function. (b) Use the 2nd derivative test to classify the critical points as relative minimums or relative

maximums. If it is not possible to classify the critical point(s) clearly explain why they cannot be classified.

(c) Use the 1st derivative test to classify the critical points as relative minimums, relative maximums or neither.

41. Given that ( )f x and ( )g x are concave down functions. If we define

( ) ( ) ( )h x f x g x= + show that ( )h x is a concave down function.

42. Given that ( )f x is a concave up function. Determine a condition on ( )g x for which

( ) ( ) ( )h x f x g x= + will be a concave up function.


Calculus I


43. For a function ( )f x determine conditions on ( )f x for which ( ) ( ) 2h x f x= will be a

concave up function. Note that there are several sets of conditions that can be used here. How many of them can you find?

The Mean Value Theorem For problems 1 – 4 determine all the number(s) c which satisfy the conclusion of Rolle’s Theorem for the given function and interval. 1. ( ) 3 24 3f x x x= − + on [ ]0,4

2. ( ) 215 2Q z z z= + − on [ ]2, 4−

3. ( ) 2 91 th t −= − e on [ ]3,3−

4. ( ) [ ]1 cosg w wπ= + on [ ]5,9

For problems 5 – 8 determine all the number(s) c which satisfy the conclusion of the Mean Value Theorem for the given function and interval. 5. ( ) 3 2 8f x x x x= − + + on [ ]3,4−

6. ( ) 3 22 7 1g t t t t= + + − on [ ]1,6

7. ( ) 2 6 3tP t t= − −e

8. ( ) ( )29 8sin xh x x= − on [ ]3, 1− −


( )5 14f = and that ( ) 10f x′ ≤ . What is the smallest possible value for ( )2f − ?

10. Suppose we know that ( )f x is continuous and differentiable on the interval [ ]6, 1− − , that

( )6 23f − = − and that ( ) 4f x′ ≥ − . What is the smallest possible value for ( )1f − ?


Calculus I



( )3 7f − = and that ( ) 17f x′ ≤ − . What is the largest possible value for ( )4f ?

12. Suppose we know that ( )f x is continuous and differentiable on the interval [ ]1,9 , that

( )9 0f = and that ( ) 8f x′ ≥ . What is the largest possible value for ( )1f ?

13. Show that ( ) 7 5 32 3 14 1f x x x x x= + + + + has exactly one real root.

14. Show that ( ) 3 26 2 4 3f x x x x= − + − has exactly one real root.

15. Show that ( ) 420 xf x x −= − e has exactly one real root.

Optimization 1. Find two positive numbers whose sum of six times one of them and the second is 250 and whose product is a maximum. 2. Find two positive numbers whose sum of twice the first and seven times the second is 600 and whose product is a maximum. 3. Let x and y be two positive numbers such the sum is 175 and ( )( )3 4x y+ + is a maximum.

4. Find two positive numbers such that the sum of one and the square of the other is 200 and whose product is a maximum. 5. Find two positive numbers whose product is 400 and such that the sum of twice the first and three times the second is a minimum. 6. Find two positive numbers whose product is 250 and such that the sum of the first and four times the second is a minimum. 7. Let x and y be two positive numbers such that ( )2 100y x + = and whose sum is a minimum.

8. Find a positive number such that the sum of the number and its reciprocal is a minimum. 9. We are going to fence in a rectangular field and have 200 feet of material to construct the fence. Determine the dimensions of the field that will enclose the maximum area.


Calculus I


10. We are going to fence in a rectangular field. The cost of material of each side is $6/ft, $9/ft, $12/ft and $14/ft respectively. If we have $1000 to buy fencing material determine the dimensions of the field that will maximize the enclosed area. 11. We are going to fence in a rectangular field that encloses 75 ft2. Determine the dimensions of the field that will require the least amount of fencing material to be used. 12. We are going to fence in a rectangular field that encloses 200 m2. If the cost of the material for of one pair of parallel sides is $3/ft and cost of the material for the other pair of parallel sides is $8/ft determine the dimensions of the field that will minimize the cost to build the fence around the field. 13. Show that a rectangle with a fixed area and minimum perimeter is a square. 14. Show that a rectangle with a fixed perimeter and a maximum area is a square. 15. We have 350 m2 of material to build a box whose base width is four times the base length. Determine the dimensions of the box that will maximize the enclosed volume. 16. We have $1000 to buy the materials to build a box whose base length is seven times the base width and has no top. If the material for the sides cost $10/cm2 and the material for the bottom cost $15/cm2 determine the dimensions of the box that will maximize the enclosed volume. 17. We want to build a box whose base length is twice the base width and the box will enclose 80 ft3. The cost of the material of the sides is $0.5/ft2 and the cost of the top/bottom is $3/ft2. Determine the dimensions of the box that will minimize the cost. 18. We want to build a box whose base is a square, has no top and will enclose 100 m3. Determine the dimensions of the box that will minimize the amount of material needed to construct the box. 19. We want to construct a cylindrical can with a bottom but no top that will have a volume of 65 in3. Determine the dimensions of the can that will minimize the amount of material needed to construct the can. 20. We want to construct a cylindrical can whose volume is 105 mm3. The material for the wall of the can costs $3/mm2, the material for the bottom of the can costs $7/mm2 and the material for the top of the can costs $2/mm2. Determine the dimensions of the can that will minimize the cost of the materials needed to construct the can. 21. We have a piece of cardboard that is 30 cm by 16 cm and we are going to cut out the corners and fold up the sides to form a box. Determine the height of the box that will give a maximum volume.


Calculus I


22. We have a piece of cardboard that is 5 in by 20 in and we are going to cut out the corners and fold up the sides to form a box. Determine the height of the box that will give a maximum volume. 23. A printer needs to make a poster that will have a total of 500 cm2 that will have 3 cm margins on the sides and 2 cm margins on the top and bottom. What dimensions of the poser will give the largest printed area? 24. A printer needs to make a poster that will have a total of 125 in2 that will have ½ inch margin on the bottom, 1 inch margin on the right, 2 inch margin on the left and 4 inch margin on the top. What dimensions of the poser will give the largest printed area?

More Optimization Problems 1. We want to construct a window whose bottom is a rectangle and the top of the window is an equilateral triangle. If we have 75 inches of framing material what are the dimensions of the window that will let in the most light? 2. We want to construct a window whose middle is a rectangle and the top and bottom of the window are equilateral triangles. If we have 4 feet of framing material what are the dimensions of the window that will let in the most light? 3. We want to construct a window whose middle is a rectangle, the top of the window is a semicircle and the bottom of the window is an equilateral triangle. If we have 1500 cm of framing material what are the dimensions of the window that will let in the most light? 4. Determine the area of the largest rectangle that can be inscribed in a circle of radius 5. 5. Determine the area of the largest rectangle whose base is on the x-axis and the top two corners lie on semicircle of radius 16. 6. Determine the area of the largest rectangle whose base is on the x-axis and the top two corners lie 24y x= − .

7. Find the point(s) on 2 2

14 36x y

+ = that are closest to ( )0,1 .

8. Find the point(s) on 2 8x y= − that are closest to ( )5,0 .

9. Find the point(s) on 22y x= − that are closest to ( )0, 3− .


Calculus I


10. A 6 ft piece of wire is cut into two pieces. One piece is bent into an equilateral triangle and the other will be bent into a rectangle with one side twice the length of the other side. Determine where, if anywhere, the wire should be cut to minimize the area enclosed by the two figures. 11. A 250 cm piece of wire is cut into two pieces. One piece is bent into an equilateral triangle and the other will be bent into circle. Determine where, if anywhere, the wire should be cut to maximize the area enclosed by the two figures. 12. A 250 cm piece of wire is cut into two pieces. One piece is bent into an equilateral triangle and the other will be bent into circle. Determine where, if anywhere, the wire should be cut to minimize the area enclosed by the two figures. 13. A 4 m piece of wire is cut into two pieces. One piece is bent into a circle and the other will be bent into a rectangle with one side three times the length of the other side. Determine where, if anywhere, the wire should be cut to maximize the area enclosed by the two figures. 14. A line through the point ( )4,1− forms a right triangle with the x-axis and y-axis in the 2nd

quadrant. Determine the equation of the line that will minimize the area of this triangle. 15. A line through the point ( )3,3 forms a right triangle with the x-axis and y-axis in the 1st

quadrant. Determine the equation of the line that will minimize the area of this triangle. 16. A piece of pipe is being carried down a hallway that is 14 feet wide. At the end of the hallway there is a right-angled turn and the hallway narrows down to 6 feet wide. What is the longest pipe (always keeping it horizontal) that can be carried around the turn in the hallway? 17. A piece of pipe is being carried down a hallway that is 9 feet wide. At the end of the hallway there is a right-angled turn and the hallway widens up to 21 feet wide. What is the longest pipe (always keeping it horizontal) that can be carried around the turn in the hallway? 18. Two poles, one 15 meters tall and one 10 meters tall, are 40 meters apart. A length of wire is attached to the top of each pole and it is staked to the ground somewhere between the two poles. Where should the wire be staked so that the minimum amount of wire is used? 19. Two poles, one 2 feet tall and one 5 feet tall, are 3 feet apart. A length of wire is attached to the top of each pole and it is staked to the ground somewhere between the two poles. Where should the wire be staked so that the minimum amount of wire is used.? 20. Two poles, one 15 meters tall and one 10 meters tall, are 40 meters apart. A length of wire is attached to the top of each pole and it is staked to the ground somewhere between the two poles. Where should the wire be staked so that the angle formed by the two pieces of wire at the stake is a maximum?


Calculus I


21. Two poles, one 34 inches tall and one 17 inches tall, are 3 feet apart. A length of wire is attached to the top of each pole and it is staked to the ground somewhere between the two poles. Where should the wire be staked so that the angle formed by the two pieces of wire at the stake is a maximum? 22. A trough for holding water is to be formed as shown in the figure below. Determine the angle θ that will maximize the amount of water that the trough can hold.

23. A trough for holding water is to be formed as shown in the figure below. Determine the angle θ that will maximize the amount of water that the trough can hold.

Indeterminate Forms and L’Hospital’s Rule Use L’Hospital’s Rule to evaluate each of the following limits.

1. 3 2

3 24

6 32lim5 4x

x xx x x→−

+ −+ +

2. 6

3lim4w

w

w

−

−→−∞ +e

e

3. ( )( )0

sin 6lim

sin 11t

tt→

4. 2

3 21

8 9lim2 5 6x

x xx x x→

+ −− − +

5. 3 2

4 3 22

7 16 12lim4 4t

t t tt t t→

− + −− +


Calculus I


6. 2

2

4 1lim3 7 4w

w ww w→−∞

− ++ −

7. 2 6

2 7lim4y

y

yyy→∞

−+ee

8. ( )

( )

2

20

2cos 4 4 2lim

sin 2 2x

x xx x x→

− −− −

9. 2 6 2

3 23

3 12lim6 9x

x xx x x

+

→−

+ −+ +

e

10. ( )

( )6

sinlim

ln 5z

zzπ

→ −

11.

2

0

0limx

x t dt

x→

∫ e

12. 2lim ln 1

3ww

w→∞

−

13. ( ) ( )0

lim ln sint

t t+→

14. 2lim

z

zz→−∞

e

15. 7lim sin

xx

x→∞

16. ( )22

0lim lnz

z z+→

17. 1

0limx

xx+→

18. 1

0limt

t tt+→

+ e


Calculus I


19. 1

2lim 3x

x xx−

→∞ − e

20. Suppose that we know that ( )f x′ is a continuous function. Use L’Hospital’s Rule to show

that,

( ) ( ) ( )0

lim2h

f x h f x hf x

h→

+ − −′=

21. Suppose that we know that ( )f x′′ is a continuous function. Use L’Hospital’s Rule to show

that,

( ) ( ) ( ) ( )20

2limh

f x h f x f x hf x

h→

+ − + −′′=

Linear Approximations For problems 1 – 4 find a linear approximation to the function at the given point. 1. ( ) ( )cos 2f x x= at x π=

2. ( ) ( )2ln 5h z z= + at 2z =

3. ( ) 2 32 9 3g x x x x= − − − at 1x = −

4. ( ) ( )sin tg t = e at 4t = −

5. Find the linear approximation to ( ) ( )sin 1h y y= + at 0y = . Use the linear approximation to

approximate the value of ( )sin 2 and ( )sin 15 . Compare the approximated values to the exact

values.

6. Find the linear approximation to ( ) 5R t t= at 32t = . Use the linear approximation to

approximate the value of 5 31 and 5 3 . Compare the approximated values to the exact values. 7. Find the linear approximation to ( ) 1 xh x −= e at 1x = . Use the linear approximation to

approximate the value of e and 4−e . Compare the approximated values to the exact values.


Calculus I


For problems 8 – 10 estimate the given value using a linear approximation and without using any kind of computational aid. 8. ( )ln 1.1

9. 8.9 10. ( )sec 0.1

Differentials For problems 1 – 5 compute the differential of the given function. 1. ( ) 6 3 23 8 9 4f x x x x x= − + − −

2. ( )2 cos 2u t t=

3. ( )cos zy = e 4. ( ) ( ) ( )sin 3 cos 1g z z z= − −

5. ( ) 4 6 xR x x −= + e

5. Compute dy and y∆ for ( )siny x= as x changes from 6 radians to 6.05 radians.

6. Compute dy and y∆ for ( )2ln 1y x= + as x changes from -2 to -2.1.

7. Compute dy and y∆ for 1

2y

x=

− as x changes from 3 to 3.02.

8. Compute dy and y∆ for 14 xy x= e as x changes from -10 to -9.99.

9. The sides of a cube are found to be 6 feet in length with a possible error of no more than 1.5 inches. What is the maximum possible error in the surface area of the cube if we use this value of the length of the side to compute the surface area?


Calculus I


10. The radius of a circle is found to be 7 cm in length with a possible error of no more than 0.04 cm. What is the maximum possible error in the area of the circle if we use this value of the radius to compute the area? 11. The radius of a sphere is found to be 22 cm in length with a possible error of no more than 0.07 cm. What is the maximum possible error in the volume of the sphere if we use this value of the radius to compute the volume? 12. The radius of a sphere is found to be ½ foot in length with a possible error of no more than 0.03 inches. What is the maximum possible error in the surface area of the sphere if we use this value of the radius to compute the surface area?

Newton’s Method

For problems 1 – 3 use Newton’s Method to determine 2x for the given function and given value

of 0x .

1. ( ) 37 8 4f x x x= − + , 0 1x = −

2. ( ) ( ) ( )cos 3 sinf x x x= − , 0 0x =

3. ( ) 2 37 xf x −= − e , 0 5x =

For problems 4 – 8 use Newton’s Method to find the root of the given equation, accurate to six decimal places, that lies in the given interval. 4. 5 6x = in [ ]1,2

5. 3 22 9 17 20 0x x x− + + = in [ ]1,1−

6. 3 43 12 4 3 0x x x− − − = in [ ]3, 1− −

7. ( )4cosx x=e in [ ]1,1−

8. 22 2 xx −= e in [ ]0,2


Calculus I


For problems 9 – 12 use Newton’s Method to find all the roots of the given equation accurate to six decimal places. 9. 3 22 5 10 4 0x x x+ − − = 10. 4 3 24 54 92 105 0x x x x+ − − + =

11. ( )232 cosx x−− =e

12. ( ) ( )ln 2cosx x=

13. Suppose that we want to find the root to 3 27 8 3 0x x x− + − = . Is it possible to use 0 4x =

as the initial point? What can you conclude about using Newton’s Method to approximate roots from this example? 14. Use the function ( ) ( ) ( )2cos sinf x x x= − for this problem.

(a) Plot the function on the interval [ ]0,9 .

(b) Use 0 4x = to find one of the roots of this function to six decimal places. Did you get the

root you expected to? (c) Use 0 5x = to find one of the roots of this function to six decimal places. Did you get the

root you expected to? (d) Use 0 6x = to find one of the roots of this function to six decimal places. Did you get the

root you expected to? (e) What can you conclude about choosing values of 0x to find roots of equations using

Newton’s Method. 15. Use 0 0x = to find one of the roots of 5 32 7 3 1 0x x x− + − = accurate to six decimal places.

Did we chose a good value of 0x for this problem?


3500,000,000 1,540,000 1450P x x x x= − + −


Calculus I


How many widgets should they try to sell in order to maximize their profit? 2. A company can produce a maximum of 25 widgets in a day. If they sell x widgets during the day then their profit, in dollars, is given by, ( ) 2 31

33000 40 11P x x x x= − + − How many widgets should they try to sell in order to maximize their profit? 3. A management company is going to build a new apartment complex. They know that if the complex contains x apartments the maintenance costs for the building, landscaping etc. will be, ( ) 2 32736 211 1

5 50 15070,000C x x x x= + − + The land they have purchased can hold a complex of at most 400 apartments. How many apartments should the complex have in order to minimize the maintenance costs? 4. The production costs of producing x widgets is given by,

( ) 90,0002000 4C x xx

= + +

If the company can produce at most 200 widgets how many should they produce to minimize the production costs? 5. The production costs, in dollars, per day of producing x widgets is given by, ( ) 2 3400 3 2 0.002C x x x x= − + + What is the marginal cost when 20x = and 75x = ? What do your answers tell you about the production costs? 6. The production costs, in dollars, per month of producing x widgets is given by,

( ) 2

8,000,00010,000 14C x xx

= + −

What is the marginal cost when 80x = and 150x = ? What do your answers tell you about the production costs? 7. The production costs, in dollars, per week of producing x widgets is given by,


Calculus I


( ) 2 365,000 4 0.2 0.00002C x x x x= + + − and the demand function for the widgets is given by, ( ) 5000 0.5p x x= − What is the marginal cost, marginal revenue and marginal profit when 2000x = and 4800x = ? What do these numbers tell you about the cost, revenue and profit? 8. The production costs, in dollars, per week of producing x widgets is given by,

( ) 2 56,000800 0.008C x xx

= + +

and the demand function for the widgets is given by, ( ) 2350 0.05 0.001p x x x= − − What is the marginal cost, marginal revenue and marginal profit when 175x = and 325x = ? What do these numbers tell you about the cost, revenue and profit?


CALCULUS I - Google Sites · PDF fileCalculus I © 2007 Paul Dawkins ii Preface Here are...

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