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Contents

1 Preface 3

2 Introduction to Calculus 4

3 Limits 6

4 The ", � Definition of a Limit 9

5 One-sided Limits 12

6 Limits at infinity (horizontal asymptotes) 13

7 Infinite limits (vertical asymptotes) 15

8 Trigonometric limits 17

9 Continuity 18

10 The Derivative at a Point 21

11 The Derivative as a Function 23

12 Di↵erentiation Rules 27

13 Rates of Change 29

14 Trigonometric Derivatives 31

15 The Chain Rule 33

16 Implicit Di↵erentiation 34

17 Related Rates 36

18 Linear Approximation and Di↵erentials 39

19 The Extreme Value Theorem 41

20 The Mean Value Theorem 44

21 Increasing and Decreasing Functions and the First Derivative Test 45

22 Concavity 48

23 Optimization 49

24 Newton’s Method 50

1

25 Antiderivatives 52

26 Area, Estimating with Finite Sums 54

27 Area with Infinite Sums 56

28 The Definite Integral 60

29 The Fundamental Theorem of Calculus 64

30 u-Substitution 67

31 The Area Between Two Curves 69

2

1 Preface

This work is based on lectures I give as an instructor in MATH 205, Calculus I, at Kapi‘olaniCommunity College (KCC) in Honolulu, Hawai‘i. The course is a requirement of all STEM (Science,Technology, Engineering, and Mathematics) degree pathways at KCC, and lasts one semester.This work was supported in part by the National Science Foundation, Hawaii’s Pre-EngineeringEducation Collaboration, Award no. 1037827. Thank you to Sunyeen Pai, who proofread, ThorChristensen, my first calculus teacher, and Herve Collin for suggesting the project.

-Austin Anderson, PhDFebruary 8, 2016

3

2 Introduction to Calculus

I have heard two descriptions of calculus that stuck with me through the years. If someone asksyou “what is calculus?” at a party, tell them one of these:

1) Calculus is the mathematics of change. (you know, velocity)2) Calculus is finding areas of weird shapes. (like under a parabola)In truth, I think calculus is both of these and their connection. The math of change pretty

much describes derivatives, and finding areas of weird shapes is the main application of integrals.Calculus is understanding both derivatives and integrals, and learning The Fundamental Theoremof Calculus, which says derivatives and integrals are inverses (like how addition and subtraction areinverses, or multiplication and division are inverses). The ancient Greeks knew something aboutderivatives and areas, but they did not know the Fundamental Theorem of Calculus. Isaac Newtonand Gottfried Wilhelm von Leibniz independently figured out “the calculus” (that is what theyused to call it) at the end of the 1600s. Newton used “the calculus” to explain gravity, as in whyapples falling o↵ trees and the moon going around the earth are the same phenomenon. Newton’spaper (more like a huge book with volumes I guess) Principia Mathematica is sometimes thoughtof as the greatest scientific achievement of humankind.

We want to learn about derivatives first, and integrals later. Derivatives are a special case ofsomething called a limit, so most textbooks start with limits and then go to derivatives. Here wewill do an example of a derivative. It is ok if it seems di�cult to you, because we will spend a lotmore time going into a lot more detail and seeing many more examples in the next two chapters.We do it here to motivate what follows, and hope that this seems easy by the end of the class.

If there is only one thing you remember from this class, it should be “the derivative is theslope of the tangent line” (or “the derivative is the instantaneous rate of change” but we willsave that for later).

Example. Find the slope of the tangent line to y =1

2x

3 at the point�1, 1

2

�.

The picture is this:

-2.4 -2 -1.6 -1.2 -0.8 -0.4 0 0.4 0.8 1.2 1.6 2 2.4

-1.6

-1.2

-0.8

-0.4

0.4

0.8

1.2

1.6

4

The black curve is the cubic function f(x) = 12

x

3, and the red line is the tangent line. Thisword “tangent line” comes from the same root as the word “tangible,” and it touches the curve at asingle point. The salient trait of the tangent line is that it goes in the same direction as the curve.The curve and line intersect at the point

�1, 1

2

�. Geometrically (visually), tangent lines are easy to

understand, but analytically (with equations) they require a solid precalculus background and thebig idea of this chapter: limits. Without limits, we cannot get tangent lines, but we can still getsecant lines. A secant (same root as “intersect”) line crosses the curve at two points, and precalculus

teaches us all about this. The slope of the line through (x1

, y

1

) and (x2

, y

2

) is m =y

2

� y1

x

2

� x1

. Think

about using x instead of x1

, x+ h instead of x2

, and f(x) instead of y. The slope formula becomes

m =f(x+ h)� f(x)

h

, which is the most convenient form for us here, and is commonly called a

di↵erence quotient. Since h is the distance between the two points on the secant line, if we shrinkh to 0 we get a tangent line at x, as the following diagram attempts to illustrate.

x (x+ h)

y

secant lines

x

y

tangent line

The above diagram is a big deal. I still remember when my calculus teacher drew that on theboard; he made us put down our pencils and just watch, something which he only did for that onemoment in the entire class. The idea is that a tangent line is a limit of secant lines as h goes to0. We have not really learned what a limit is yet, so its ok to be wondering as we begin using thelanguage of calculus.

Back to f(x) = 12

x

3, we are going to calculate the slope of the secant lines. Your precalculustraining should enable you to follow the next display. (It often takes a long time, thinking andchecking with scratch paper, to follow a calculation. You may need to fill in steps on your own. Ifyou are stuck after spending a long time on it, ask for help.)

5

f(x+ h)� f(x)h

=1

2

(x+ h)3 � 12

x

3

h

(1)

=1

2

(x3 + 3x2h+ 3xh2 + h3)� 12

x

3

h

(2)

=1

2

x

3 + 32

x

2

h+ 32

xh

2 + 12

h

3 � 12

x

3

h

(3)

=3

2

x

2

h+ 32

xh

2 + 12

h

3

h

(4)

=h

�3

2

x

2 + 32

xh+ 12

h

2

�

h

(5)

=3

2x

2 +3

2xh+

1

2h

2 for h 6= 0. (6)

The first step is using the given function. (2) is an algebra exercise, like the FOIL method. (5)comes from simple factoring. In (6), it is important to note that you cannot cancel the h unlessh 6= 0, because 0

0

is undefined.So, 3

2

x

2 + 32

xh+ 12

h

2 is the slope of a secant line to y = 12

x

3. To find the tangent line in the firstpicture, we let x = 1 and we take a limit as h! 0. It amounts to simply plugging in x = 1, h = 0,giving the slope 3

2

(1)2 + 32

(1)(0) + 12

(0)2 = 32

.PAU. The answer is 3

2

. Look carefully at the first picture, and try to measure the rise over therun (slope) of the red line, a.k.a. the tangent line. The slope should look like 3

2

. Later, we will saythe derivative of f(x) = 1

2

x

3 at 1 is 32

, and learn a very quick way to calculate it.

3 Limits

Our main goal is to understand how to find tangent lines to graphs of functions. When we look

at the slope formulaf(x+ h)� f(x)

h

for a secant line, and then let h shrink to 0, we are taking

a limit. This is probably the most important limit, but it is not the easiest to understand. Somelimits are easier, and limits are important in and of themselves. In this section we look at someexamples of limits.

limx!a

f(x) = L means f(x) approaches L as x approaches, but does not equal, a.

The notation limx!a

f(x) = L is standard and every detail matters; we will use it frequently in

all that follows. The word approaches is the best we can do here, but is not really satisfactoryto a rigorous mathematician. I will often say “gets infinitely close to” in place of “approaches.”In calculus, we need to talk about infinity a lot. Infinitely large things are just called “infinite,”and infinitely small (technically “infinitesimally small”) things are called “infinitesimal.” When xapproaches a number a, the distance |x� a| becomes infinitesimally small, and the function values(the y-values) must get infinitely close (so the distance |f(x) � L| gets infinitesimally small) to L

6

for the limit to exist. This is what makes calculus conceptually di�cult and what gave the Greekstrouble (google Zeno’s paradoxes). In fact, even Isaac Newton did not do a great job at explaininglimits. Newton used an idea that he called fluxions, which we do not use today because we havefound something better. Modern calculus usually uses the "-� (epsilon-delta) definition of a limitattributed to the mathematician Karl Weierstrass in the 1800s. The "-� definition is covered insection 4. For now, we will continue without probing the matter too deeply, and be content withvocabulary like “approaches,” “gets close to,” and “goes to,” or just an arrow, !.

Example. limx!3

x� 3x� 3

First, focus on the function f(x) =x� 3x� 3. The domain is all real numbers except 3. If you plug

in any number in the domain, the top and bottom cancel, giving 1. I.e.,

f(x) =x� 3x� 3 = 1 for x 6= 3. (7)

It is important to emphasize that the equality is only true when x 6= 3, because 0/0 is undefined.The graph of f looks like this:

x

y

(3, 1)

Note the puka (hole) at (3, 1). The

function is undefined at x = 3, so we can’t plot a point there. The limit limx!3

x� 3x� 3 describes

the y-value of the graph as x approaches, but does not equal, 3. Since the y-values near x = 3 are

all 1, we write limx!3

x� 3x� 3 = 1, and say “the limit of

x� 3x� 3 as x goes to 3 is 1.”

Example. limx!1

x

4 � 1x� 1

We will try to figure out this limit without a graph. Our goal is to determine what the y-valuesof the function y = f(x) = x

4�1x�1 approach as x approaches 1. Note that f(1) is undefined, but that

7

does not mean we can’t find the limit. Since 1 is the only x value we cannot plug in to f(x), wetry to get infinitely close to 1 without touching it. Of course, we can’t really do anything infinitein our finite lifetime, so we just go for a little while and hope to see a pattern. The table of valueshere is nothing more than plugging in x values, using a calculator to get the y values, and roundingto a few decimal places.

x 0.9 0.99 0.999 1.001 1.01 1.1f(x) 3.439 3.9404 3.994 4.006 4.0604 4.641

Note that 0.9, 0.99, 0.999 is chosen to approach 1, and we figure that 3.439, 3.9404, 3.994 is

approaching 4. Therefore, limx!1

x

4 � 1x� 1 = 4.

We have evaluated the limit and gotten the correct answer, but we are going to do it again, adi↵erent way. Tables are good for understanding, but bad for taking a lot of time and not alwaysworking. With some algebra and logic we can do a faster, better job evaluating limits. The keyfor lim

x!1x

4�1x�1 is to factor the numerator. Although there is more than one way to do this, the

following works and can be checked by multiplying:

x

4 � 1x� 1 =

(x� 1)(x3 + x2 + x+ 1)x� 1 = x

3 + x2 + x+ 1 for x 6= 1.

(Be careful; it is incorrect to leave out “for x 6= 1”.) We deduce that f(x) is the same as x3+x2+x+1everywhere except 1, and in making our table of values we could have used the simpler formula. Itis obvious⇤ that x3 + x2 + x+ 1 is going to get closer to 13 + 12 + 1+ 1 = 4 as x gets closer to 1, sowe know that lim

x!1(x3 + x2 + x + 1) = 4. Putting it all together, here is what we write to evaluate

the limit:

limx!1

x

4 � 1x� 1 = limx!1

(x� 1)(x3 + x2 + x+ 1)x� 1

= limx!1

(x3 + x2 + x+ 1)

= 13 + 12 + 1 + 1

= 4.

Note carefully that we need the limit notation in the first 3 expressions, and it goes away whenwe plug in 1. Doing this incorrectly can lose you points on the test, because the correct notationdemonstrates understanding.

CORRECT: limx!1

(x� 1)(x3 + x2 + x+ 1)x� 1 = limx!1(x

3 + x2 + x+ 1)

WRONG:(x� 1)(x3 + x2 + x+ 1)

x� 1 = (x3 + x2 + x+ 1)

The latter is wrong because those two expressions are not always equal, which is shown byplugging in x = 1 to each expression and getting a false statement. (Undefined = 4 is false.)However, when we have the limit notation out front it is ok, because the limit notation indicates xapproaches but does not equal 1.

⇤ “Obvious” is a risky word to use in math. Used here, we are glossing over a big part ofunderstanding limits, namely, the limit laws:

8

The limit of a constant is the constant:

limx!a

k = k, for a constant k.

As x approaches a, x approaches a :

limx!a

x = a (duh, sorta).

Constant Multiple Law:

limx!a

[kf(x)] = k limx!a

f(x), for a constant k.

Sum Law:

limx!a

[f(x) + g(x)] = limx!a

f(x) + limx!a

g(x), provided the limits exist.

Product Law:

limx!a

[f(x)g(x)] =⇣limx!a

f(x)⌘⇣

limx!a

g(x)⌘, provided the limits exist.

There are more, including the Di↵erence Law, Quotient Law, and Power Law. The limit lawsare especially important in proofs, but teaching a rigorous understanding of proofs is typicallypostponed until a student has completed their first calculus courses. (My favorite class in which tolearn and use proofs is linear algebra, which at UH Mānoa is MATH 311.)

4 The ", � Definition of a Limit

In the previous section, we used the word “approach” in defining a limit. In advanced mathematics,this doesn’t cut it. In this section, we will go into more detail about the definition of a limit.However, some calculus classes will skip this section because it is conceptually di�cult. By skippingit, the beginning student loses little, and can still understand everything that follows. When I wasa student, I didn’t really learn the ideas in this section until a 300 level math class, long after Ihad finished my first calculus series (at UH the first calculus series is Calc I - IV). In that 300 levelclass, called Real Analysis, one typically learns the definition of a real number as an equivalenceclass of Cauchy sequences. It addresses problems that can pretty much be avoided in beginningcalculus. If one focuses on the details of real numbers, slightly strange things happen, such as thefact that a real number does not have a unique decimal expansion. 0.999999999.... = 1 (woah).In my experience, that is the first place in a mathematics education where the ", � (say “epsilon,delta”) definition of a limit is truly necessary. Also, once a student has learned quite a bit aboutmathematical proof, they appreciate the usefulness of the ", � definition of a limit, and can workwith the quantifiers more easily.

Thus, I personally prefer to learn about proofs, quantifiers, and the definition of a real numberbefore the ", � definition of a limit. However, Calc I books typically include it, and at Mānoa youmight be expected to cover it in Calc I. When I was a teaching assistant in graduate school, aMānoa professor told me that the main reason he insisted on covering the ", � definition in Calc Iwas so students could learn to use quantifiers, so let’s start with that.

9

“Quantify” means count, or measure with a number. If you “quantify” your success, you givea number for how much money you have, your G.P.A., how many years of school you finished, orsomething like that. Clearly a quantifier quantifies something. Some examples of quantifiers yousee in the language of mathematics are:

for all ; It means none are left out.there exists some; It means there is at least one.there exists a unique; It means there is one and only one.You might see simple quantifiers in MATH 100 or a philosophy class that covers logic. For

example:“All girls are people.”“Some people are girls.”“One person is typing this sentence.”“There exists no person who is not a person.” (The quantity is 0.)These are logically valid statements, silly or obvious as they may sound. Math is all about logic

validity, and it’s not always obvious. Careful language is important. Another important idea inlogic and proof is that of implication.

You are a girl implies you are a person.You are a person does not imply you are a girl. (Because you could be a boy or, you know, a

non-girl person.)ANYWAY, below is the most popular definition of a limit in modern math (there are others less

popular), attributed to Karl Weierstrass in the 1800s.

limx!a

f(x) = L

means that for all " > 0, there exists some � > 0 such that

0 < |x� a| < � implies |f(x)� L| < ".

Let’s attempt to dissect the definition and put it in layman’s terms. (Don’t get your hopes up,though; it’s written as well as it can be–that’s why we use it. The layman’s terms were already insection 3.) In section 3 we said “as x approaches, but does not equal, a.” This is represented bythe 0 < |x � a| < � above. Think of " and � as very small numbers. The expression |x � a| is thedistance between x and a, so |x�a| < � means x and a are very close. If |x�a| = 0 then x = a, butwe don’t allow that in a limit, so 0 < |x� a| is required. The expression |f(x)� L| < " means they-values f(x) are very close to the limit L. How close? Within ", an arbitrarily tiny distance. Wesay " is arbitrary because the statement holds for all " > 0. That’s a strong statement. It needsto be true for " = 0.1, " = 0.00000000000001, ..., anything! (positive, not 0) So lim

x!a f(x) = Lmeans given any " > 0, we can choose a nonzero � that ensures f(x) is within " of L wheneverx 6= a is within � of a.

Example: Suppose f(x) = 2x+ 1, a = 3, and " = 0.1. Show that there exists some � > 0 suchthat 0 < |x� a| < � implies |f(x)� 7| < ".

10

We know by common sense (section 3) that

limx!3

f(x) = 2(3) + 1 = 7.

So the limit L = 7, and in this example we are working on verifying the statement limx!3 f(x) = 7

using the formal definition of a limit. For this function and " = 0.1, we can choose � = 0.05 (oranything smaller) to make the implication in the definition true.

0 < |x� 3| < � (1)) 0 < |x� 3| < 0.05 (2)) |x� 3| < 0.05 (3)) �0.05 < x� 3 < 0.05 (4)) �0.1 < 2(x� 3) < 0.1 (5)) �0.1 < 2x+ 1� 7 < 0.1 (6)) |2x+ 1� 7| < 0.1 (7)) |f(x)� 7| < " (8)

The chain of implications () means “implies”) can be summarized by the statement that wewant: there exists some � > 0 such that 0 < |x � 3| < � ) |f(x) � 7| < ". Let’s talk about eachstep. (1) is the assumption we are starting with, and (2) is identical with � plugged in. We chose� = 0.05 but there are others that work. “There exists some” is the quantifier, so one is enough. (3)is not needed, but in this case the function is defined at 3, so x = a is not a problem. The reason Iincluded (3) is that (3) ) (4) is an exact statement that I teach in College Algebra, MATH 103: if|b| < c for some positive constant c, then �c < b < c. (5) comes from multiplying by 2. (6) is basicalgebra since 2(x� 3) = 2x� 6 = 2x + 1� 7, but it is done with (8) in mind specifically, because(7) is the MATH 103 fact again and (8) is plugging in the given f and ".

Example: Prove that limx!3(2x+ 1) = 7 using the ", � definition of a limit.

Now we have to do it for all ". The � that we choose depends on ". For this function, we willsee that � = "

2

works.

Proof. Given " > 0, let � ="

2. Then

0 < |x� 3| < � (1)

) 0 < |x� 3| < "2

(2)

) |x� 3| < "2

(3)

) �"2< x� 3 < "

2(4)

) �" < 2(x� 3) < " (5)) �" < 2x+ 1� 7 < " (6)) |2x+ 1� 7| < " (7)) |(2x+ 1)� 7| < ". (8)

11

(Same steps as the first example, but now with arbitrary ".)Therefore, lim

x!3(2x+ 1) = 7.

5 One-sided Limits

x 0.9 0.99 0.999 1.001 1.01 1.1f(x) 3.439 3.9404 3.994 4.006 4.0604 4.641The above table, from section 3, shows values of a function f such that lim

x!1 f(x) = 4. Thevalues 3.439, 3.9404, 3.994 are y-values at points where x < 1, and pertain to a one-sided limit,namely the left-hand limit. For the left-hand limit, we use a little minus sign superscript in ourlimit notation, and write lim

x!1�f(x) = 4.

limx!a�

f(x) = L means f(x)! L as x! a for x < a.limx!a+

f(x) = L means f(x)! L as x! a for x > a.

The way I remember it is that 1� is for x-values like 0.9, 0.99, 0.999, etc.... The x values for 1�

are 1 minus a tiny number. On the other side of 1 in the table, the y-values 4.641, 4.0604, and 4.006are approaching the right-hand limit, lim

x!1+ f(x) = 4. In this case, both one-sided limits are 4,so the “regular” limit is 4. Sometimes one or both of the one-sided limits do not exist. Sometimesthey both exist, but are not equal, in which case the regular limit does not exist.

Example.y =x� 3|x� 3| .

Below is a graph of f(x) =x� 3|x� 3| . Plot points carefully to verify that you understand the

graph–you might be expected to draw it on your own on a test.

x

y

(3, 1)

(3,�1)

The y-values are 1 for x > 3 and �1 for x < 3. We can see from the picture that the right-handlimit is 1 and the left-hand limit is �1. We write

12

limx!3+

f(x) = 1, for the right-hand limit, and

limx!3�

f(x) = �1, for the left-hand limit.Since the one-sided limits are not equal, lim

x!3f(x) does not exist. Make sure you remember that

the superscript + is for the right-hand limit, and it means from the right (“from” not “to”).Example. Graph of y = f(x) below.

x

y

(3,�1)

The limits are limx!3+

f(x) = 0, limx!3�

f(x) = �1, and limx!3

f(x) d.n.e..

Students need to learn the importance of “from” in the concept of one-sided limits. If you lookat limits “to” the right and left, you are looking at something di↵erent, called limits at infinity.

6 Limits at infinity (horizontal asymptotes)

As we head o↵ a graph to the right, x goes to infinity. To the left, x goes to �1.

limx!1

f(x) = L means f(x)! L as x increases without bound.lim

x!�1f(x) = L means f(x)! L as x decreases without bound.

When the y-values approach a limit as x! ±1, the graph has a horizontal asymptote. Limits atinfinity and horizontal asymptotes are basically the same concept, so review the following examplesfrom precalculus.

Examples.

(a) y =3

x

2 + 1.

The degree of the numerator is 0, and the degree of the denominator is 2. The horizontal asymp-

tote is y = 0 (the x-axis). Try to remember “big bottom goes to 0.” Therefore limx!1

3

x

2 + 1= 0.

A table of values confirms this; the graph also.

13

x 1 10 100 1000 10000 to 1y 1.5 0.029703 0.00029997 0.000003 0.00000003 to 0

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

Figure 1: y =3

x

2 + 1

(b) y =3x� 32x+ 2

.

When the top and bottom have the same degree, the lead coe�cients determine the horizontal

asymptote. This function has y =3

2as its horizontal asymptote. For rational functions, the

asymptote goes both left and right, so here

limx!1

3x� 32x+ 2

=3

2and lim

x!�1

3x� 32x+ 2

=3

2,

as the graph demonstrates.

-10 -7.5 -5 -2.5 0 2.5 5 7.5 10

-5

-2.5

2.5

5

Figure 2: y = 1.5 asymptote

(c) y =�x2 + 4x+ 1

.

The top is bigger, and there is no horizontal asymptote. In precalculus you learn to use divisionto find oblique asymptotes when the degree of the numerator is only 1 greater than the degree

14

of the denominator, but the most we can say about limits in this case is

limx!1

�x2 + 4x+ 1

= �1 and limx!�1

�x2 + 4x+ 1

=1.

-20 -15 -10 -5 0 5 10 15 20

-10

-5

5

10

Figure 3: no horizontal asymptote

The 3 examples above demonstrate the 3 possibilities for limits at infinity of rational functions,which are functions of the form p/q where p and q are polynomials. Rational functions are themost common functions with asymptotes that you study. However, there are other interestingfunctions with horizontal asymptotes. Some of these other functions have a di↵erent horizontalasymptote to the left than to the right. My favorite one of these is the inverse tangent function,a.k.a. f(x) = arctan x, whose graph is here:

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

Figure 4: y = arctan x

limx!1

arctan x =⇡

2and lim

x!�1arctan x = �⇡

2.

7 Infinite limits (vertical asymptotes)

You shouldn’t be surprised, since we just reviewed horizontal asymptotes (h.a.s), that now youshould review vertical asymptotes (v.a.s). For horizontal, x ! ±1, and for vertical, y ! ±1.

15

Here is an example from trigonometry, y = tan x, whose inverse we just talked about in the lastsection.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

Figure 5: y = tan x

Here limx!⇡2

�tan x =1 and lim

x!⇡2+tan x = �1.

The graph above has vertical asymptotes x = ⇡/2 + ⇡k for k = 0,±1,±2, ..., which are also thezeroes of cos x. Note that x = ⇡

2

is the equation of a vertical line, and you need to write the “x =”part in your work for indicating v.a.s, as well as “y =” when indicating h.a.s. Draw the vertical linex = ⇡/2 ⇡ 1.6 on the graph above. Your eye should follow the graph up at x = ⇡/2 from the left toget lim

x!⇡2� tan x =1, and down from the right to get lim

x!⇡2+ tan x = �1. Since the one-sided

limits are not equal, all we can say about the regular limit is limx!⇡2 tan x d.n.e..

Usually, when a function is a quotient, like a rational function or tan x = sinxcosx

, the verticalasymptotes happen where the denominator is 0. I say “usually”, not always, because of exampleslike f(x) = x�3

x�3 , where the bottom is 0 at x = 3, but the graph has no vertical asymptote.

When the bottom goes to 0 and the top does not, you get an infinite limit, vertical asymptote.

A more precise theorem than the framed statement is typically included in calculus textbooks,but I wrote it colloquially so it is easier to remember. It is the final piece of the flow chart forevaluating limits of elementary functions I will include later. Technically it is correct to say “doesnot exist” for all infinite limits, because y !1 means y does not approach a real number. However,it is best to indicate the sign if you can.

Examples.

(a) y =3

x

2 + 1.

There is no vertical asymptote, because the bottom is never 0 in the real numbers. This functionhas no infinite limits, which is di↵erent from saying “no limits at infinity.”

(b) y =3x� 32x+ 2

.

The vertical asymptote is x = �1, since for x = �1 the bottom is 0 and the top is not.

16

-10 -7.5 -5 -2.5 0 2.5 5 7.5 10

-5

-2.5

2.5

5

Figure 6: x = �1 asymptote

The infinite limits are limx!�1�

3x� 32x+ 2

= 1 and limx!�1+

3x� 32x+ 2

= �1. You can get this by thegraph, or by a table of values. I would say the quickest way to figure out a problem like thisis to think of a table of values, and you can almost do it in your head. For the left limit asx! �1�, I imagine plugging in �1.1. The factored form 3(x�1)

2(x+1)

is clearly a negative divided by

a negative which is positive, so we get positive infinity as the limit, limx!�1�

3x� 32x+ 2

=1.

Recap: limx!�1�

3x� 32x+ 2

=1.

Once I see that the bottom is 0 and the top isn’t, I know we are dealing with an infinite limit.The work I might show is

3x� 32x+ 2

=3(x� 1)2(x+ 1)

. Think3(�1.1� 1)2(�1.1 + 1) =

(�)(�) = +)1 :)

Of course, it is not standard to have minus signs by themselves in a mathematical expression,but it makes sense here, because the sign is all you care about once you know it is an infinitelimit.

For the right-hand limit limx!�1+

3x� 32x+ 2

, think3(�0.9� 1)2(�0.9 + 1) =

�+

= � ) �1, so limx!�1+

3x� 32x+ 2

=

�1.

8 Trigonometric limits

The most important trig function is f(x) = sin x, and the most important aspect of sinx in termsof calculus is the cyclical nature of its derivatives. It’s too soon to learn that the derivative of sin xis cos x (whoops, spoiler!), but we will need the following limits to prove it later.

The two important trig limit facts:

lim✓!0

sin ✓

✓

= 1 and lim✓!0

1� cos ✓✓

= 0. (✓ in radians)

17

These are facts I will not explain here. Understanding these limits requires a solid geometricunderstanding of the definition of the sine and cosine functions. A proof uses geometry and theSqueeze Theorem for limits. You should look up the proof if you are interested. Engineering andphysics students use the sine enough to make it worthwhile, but business and life science studentswill probably be fine in their fields without it. Be careful with trigonometric limits, because somehowtheir nature allows students to trick themselves into thinking they understand when they actuallydo not. I think its the notation. Review things like sin2 x means (sin x)2 which is NOT sin(x2), andtan�1 x = arctan x which is NOT cot x = 1

tanx

.

Example: limx!0

sin 8x

x

.

Lucky Larry gets this limit right for the wrong reasons. Note that sin 8x means sin(8x), and theorder of operations for evaluation are multiply by 8, then take the sine. That is totally di↵erentfrom taking the sine then multiplying by 8 (check sin(8 · 30�) 6= 8 sin(30�)). You CANNOT “factorout” an 8 from inside a function, especially the sine function. To evaluate the limit, you need towrite something like the following:

limx!0

sin 8x

x

= limx!0

✓sin 8x

x

· 88

◆(1)

= limx!0

8 sin 8x

8x(2)

= limu!0

8 sin u

u

(3)

= limu!0

✓8 · sin u

u

◆(4)

= 8

✓limu!0

sin u

u

◆(5)

= 8 · 1 = 8. (6)

You can convince me you understand this without writing quite as much as I did above, but Iwant to be clear about the rules. (1) is merely multiplying the given function by 1, which is a classicalgebra technique. (2) is arithmetic; multiply the fractions and put the constants in front. (3) is asubstitution, u = 8x, which emphasizes the form in the important trig limit fact that you will use.

In the important trig limit fact lim✓!0

sin ✓

✓

= 1, the angle inside the sine and the denominator must

match exactly. Of course, ✓ is a dummy variable, and the fact is just as true if ✓ is a u instead.That is how we got (6). The substitution in (3) is also tricky because you have to understand thatif x! 0, then 8x! 0 also. Note that (5) comes from the constant multiple limit law.

9 Continuity

The idea of a continuous function is pretty intuitive. A continuous function has no holes or breaks.You can draw it without lifting your pencil o↵ the page. However, modern functions can beimpossible to draw (google the Dirichlet function or Thomae’s function), so the intuition breaksdown sometimes. The mathematical definition of continuity is simple and clear.

18

Definition. A function f(x) is continuous at x = c if limx!c

f(x) = f(c).

Note that the equation in the definition is false if either f(c) does not exist or the limit does notexist, so in those cases the function is not continuous at c. Pictures are valuable for the concept ofcontinuity.

-10 -7.5 -5 -2.5 0 2.5 5 7.5 10

-5

-2.5

2.5

5

-10 -7.5 -5 -2.5 0 2.5 5 7.5 10

-5

-2.5

2.5

5

Figure 7: Polynomials and sine waves are continuous at all x.

x

y

(3, 1)

Figure 8: Here f(3) is undefined, so f is not continuous at x = 3.

19

x

y

(3, 1)

Figure 9: Here, f(3) = 2 6= 1 = limx!3

f(x), so f is not continuous at x = 3.

x

y

(3,�1)

Figure 10: Here limx!3

f(x) d.n.e. (the one-sided limits are not equal), so f is not continuous at 3.

For the most part, STEM majors will be dealing with everywhere-continuous functions, or atleast functions that are continuous on their domains. Continuity becomes increasingly important inmath classes beyond Calc 1, because it plays a big role in the theoretical framework of calculus andthe real numbers. However, all of the basic functions you learn about in precalculus are continuouson their domains, except piecewise functions, which are specifically designed to explore the conceptof continuity. Rational functions are not continuous at vertical asymptotes, and radicals like

px

are continuous at every point interior to their domain and have one-sided continuity at points onthe boundary of their domain. When I say “if you can plug in, do it” when evaluating a limit, it is

20

because the functions involved are continuous. In fact, that is exactly what continuity tells us, i.e.,it tells us when you can plug in to find a limit. Once again, by definition, f is continuous at x = cif lim

x!cf(x) = f(c).

When you replace regular limits with one-sided limits in the definition of continuity, you getone-sided continuity. In the figure above, lim

x!3�f(x) = �1 and lim

x!3+f(x) = 0. Note that f(3) = 0.

x

y

(3,�1)

Figure 11: Here f is continuous from the right at 3.

Therefore, limx!3+

f(x) = f(3), meaning f(x) is continuous from the right at x = 3. Since �1 6= 0, weknow lim

x!3�f(x) 6= f(3), so f is not continuous from the left at x = 3. The only discontinuity is at

x = 3, so we can list intervals where f is continuous. We write “f is continuous on (�1, 3) and on[3,1).” Note the bracket vs parenthesis. Writing “f is continuous on [3,1)” does not mean that fis continuous at 3. This confuses students, because 3 is in that interval. We do this because we wantto indicate one-sided continuity, and the notation is convenient. When we write “f is continuouson [a, b],” we do not mean at a or at b. Rather, we mean that f is continuous at every point in theopen interval (a, b), f is continuous from the left at b, and f is continuous from the right at a.

10 The Derivative at a Point

Finally, we get to the main reason we learned about limits in the first place. This brings us fullcircle, so look back at section 2 before reading this.

Definition: The derivative of a function f(x) at x = x0

is denoted f 0(x0

) and defined bythe limit

limh!0

f(x0

+ h)� f(x0

)

h

= f 0(x0

).

21

Once again, if you only remember one thing from this class, it should be that the derivative isthe slope of the tangent line to a curve at a point. The idea is that the tangent line slope is a limitof secant line slopes. This section focuses on using the definition of the derivative to find tangentlines. Now that we are good at evaluating limits, we can do this more easily than we did in section2. In the next chapter, we are going to learn rules that make it easier still (yay!). Students whohave taken calculus before probably remember that the derivative of x3 is 3x2 (and the easy trickto deduce that, called the “power rule”), but right now we are calculating derivatives from firstprinciples, i.e., the definition.

Example. Find the derivative of f(x) = x2 + x at x0

= 1.Note that this example could have said “find the slope of the tangent to y = x2 + x at x

0

= 1,”which means the same thing. We simply plug everything in to the definition. Evaluating thesederivatives is one of the main goals for which precalculus classes are preparing you, so you need torely heavily on algebra and function notation skills, in addition to your newly learned understandingof limits. We get

f

0(x0

) = limh!0

f(x0

+ h)� f(x0

)

h

(1)

= limh!0

f(1 + h)� f(1)h

(2)

= limh!0

(1 + h)2 + (1 + h)� (12 + 1)h

(3)

= limh!0

1 + 2h+ h2 + 1 + h� 2h

(4)

= limh!0

3h+ h2

h

(5)

= limh!0

h(3 + h)

h

(6)

= limh!0

(3 + h) (7)

= 3 + 0 = 3. (8)

In the above display, (2) comes from plugging in the given x0

. Use function notation for the givenfunction to get (3), which may be the hardest step for students who are not strong in precalculus.(4)-(6) are from elementary algebra (FOIL out (1 + h)2 = (1 + h)(1 + h) so you do not fall for theFreshman’s Dream!), and (7) is where we need limit notation to emphasize h! 0 but h 6= 0. (8) isevaluating the limit; hence the limit notation goes away, and the answer is 3.

The graph shows the tangent line, and you can see that the slope is 3. The equation of thetangent line in point slope form is

y � 2 = 3(x� 1).

3 is the slope, the 1 comes from x0

and the 2 comes from the given function f , since f(1) = 12+1 = 2.

22

-3 -2 -1 0 1 2 3 4 5 6

-1

1

2

3

4

Figure 12: y = x2 + x and its tangent line at x0

= 1

11 The Derivative as a Function

All we do in this section is replace the fixed number x0

in section 10 by the variable x. It seemssimple, but it is such a key step toward our goals that we devote an entire section to it.

Definition: The derivative of a function f(x) is another function, and is denoted f 0(x) anddefined by the limit

limh!0

f(x+ h)� f(x)h

= f 0(x),

provided the limit exists.

Example f(x) = �x2. Find f 0(x) and f 0(1), f 0(2), f 0(�2), f 0(�0.5), f 0(0), f 0(⇡), f 0(e), and f 0(3.14).

The silly list of f 0 values is just to scare students away from plugging in numbers too early. Insection 10 we plugged in x

0

whenever we wanted, but from now on we won’t do it until the veryend. It emphasizes that f 0 is a function of x. Once you calculate f 0(x), you can find the slope ofthe tangent line at loads of di↵erent points just by plugging into the formula for f 0. Other thanthat, the calculations are exactly the same as in section 10. We once again reinforce our algebraand precalculus skills by working out

23

f

0(x) = limh!0

f(x+ h)� f(x)h

(1)

= limh!0

�(x+ h)2 � (�x2)h

(2)

= limh!0

�(x2 + 2xh+ h2) + x2

h

(3)

= limh!0

�2xh� h2

h

(4)

= limh!0

h(�2x� h)h

(5)

= limh!0

(�2x� h) (6)

= �2x� 0 = �2x. (7)

So f(x) = �2x. So f 0(1) = �2, f 0(2) = �4, f 0(�2) = 4, f 0(�0.5) = 1, f 0(0) = 0, f 0(⇡) = �2⇡,f

0(e) = �2e, and f 0(3.14) = �6.28.Below is a graph of f and all the tangent lines whose slopes we just calculated.

Suppose we wanted to graph y = f 0(x). First o↵, don’t get y = f(x) confused with y = f 0(x),because it’s not the same y. We will practice sketching f 0 based on the graph of f without evenknowing the formula for f . The key is that the slopes of f are the y-values of f 0. You have to beable to judge a lines slope just by “eyeballing” it. Steep decreasing slopes are negative numbers less

24

than �1, shallow decreasing slopes are between �1 and 0, horizontal slopes are 0, and increasingslopes are positive. When we talk about decreasing and increasing, we always read the graph leftto right. Look at the graph of f below, and try to plot the slopes as y-values at each x, therebygraphing f 0. (Don’t look at the next page until you have tried to sketch f 0.)

x

y

25

What we want to do is sketch little tangent lines on the graph of f , and then plot the slopes asy-values for f 0. The function f graphed above is decreasing at first, then constant, then becomescurved. The tangent line to a line is the line, so for the straight parts of f the y-values of f 0 areconstant, and f 0 is a flat line. Note the discontinuities of f 0 at the x-coordinates where f has acorner. The derivative is always undefined at a corner of the function. For the curved part, it ishard to graph f 0 accurately, but we plot some slope values and then connect the dots.

x

y

Figure 13: slopes (on the graph of f)

x

y

Figure 14: y-values (on the graph of f 0)

Is the graph in figure 14 what you expected when you tried to graph f 0 on the previous page?If not, try again without looking;)

26

12 Di↵erentiation Rules

Ok, so we know the definition of the derivative by heart and are good at finding limits of di↵erencequotients, i.e., derivatives. Still, the limit and di↵erence quotient take a lot of time to work out foreach function, and once we have done it many times we start to see patterns. Crazily, the patternsare very strong, and it does not take too much e↵ort to write down some rules that will makedi↵erentiation (the process of taking the derivative) waaaaay easier. The di↵erentiation operator

is denotedd

dx

, andd

dx

f(x) = f 0(x). We will use both notations (the prime and the ddx

) frequently

from now on. Why both? One reason is that a little 0 is not emphatic enough sometimes. Anotherreason is that d

dx

has the advantage of making the variable x obvious, and there are even otherreasons that we will see as we go on (there is a reason the di↵erentiation operator looks like afraction). For example, if f(x) = x, then f 0(x) = 1. However, we never write x0 = 1. (Never!)The prime is only for functions, and sometimes we want a variable besides x. Instead, we writed

dx

x = 1. Here, the x is a dummy variable, so we could also writed

dy

y = 1,d

du

u = 1,d

d✓

✓ = 1, or

even . Believe it or not, the concept of dummy variables is important, and we willsee it a lot. Note that the variable in the di↵erentiation operator might not be same as the variablein the expression that follows, in which case you have to worry about whether the expression isa constant or a nonconstant function. The derivative of a constant is 0, and the derivative of afunction depends, of course, on the function. The following rules come from the definition of thederivative, and we tend to memorize these. If we ever forget the rules or run into something new,we go back to the definition (boxed in lecture 3.1).

Di↵erentiation Rules Here f and g are di↵erentiable functions (i.e., their derivatives exist)of x.

d

dx

c = 0. The derivative of a constant is 0.

d

dx

(mx+ b) = m. The derivative of a line is its slope.

d

dx

[cf(x))] = cd

dx

f(x) = cf 0(x). This is the Constant Multiple Rule.

d

dx

x

n = nxn�1; the Power Rule (everyone’s favorite).

d

dx

[f(x) + g(x)] =d

dx

f(x) +d

dx

g(x) = f 0(x) + g0(x); the Sum Rule.

d

dx

[f(x)g(x)] = f 0(x)g(x) + g0(x)f(x); the Product Rule.

d

dx

f(x)

g(x)

�=

f

0(x)g(x)� g0(x)f(x)g

2(x); the Quotient Rule.

27

d

dx

(f � g)(x) = ddx

[f(g(x))] = f 0(g(x))g0(x); the Chain Rule.

Note that the second rule supersedes the first, because constant functions are just lines withslope 0. Combining the Constant Multiple Rule for c = �1 with the Sum Rule gives the Di↵erenceRule d

dx

[f(x)� g(x)] = f 0(x)� g0(x). Note that the Product Rule and Quotient Rule are far fromobvious, as they are not the natural thing for a newbie to expect. Review function compositionfrom precalculus before you try to learn the Chain Rule. The main goal of this chapter is to beable to di↵erentiate any function you can write down. You will become fluent in the language ofdi↵erentiation, and these rules are the grammar.

Examples: To gain learning from these examples, you need to work them all out on your own.Try to do each one before looking at the answer. You may need to fill in steps or check the algebrawith scratch work, so go get a pencil and paper before you read what follows!

Di↵erentiate the given functions.

1. s(t) = t9 � 8t2 + t� 70Finding the derivative of a polynomial is amazingly easy because of the Power Rule (the PowerRule is powerful!). Students will be able to jump straight to the answer with practice, buthere I will show some work to point out the rules we use.

s

0(t) =d

dt

(t9 � 8t2 + t� 70)

=d

dt

t

9 � ddt

8t2 +d

dt

t� ddt

70 (Sum/Di↵. Rules)

=d

dt

t

9 � 8 ddt

t

2 +d

dt

t� ddt

70 (Const. Mult. Rule)

= 9t8 � 8(2t) + 2(1)� 0 (Power Rule)= 9t8 � 16t+ 2.

2. f(t) =pt� 1

3pt

Even more amazing than its simplicity, the Power Rule works for fractional and negativeexponents! You will appreciate this more if you did several calculations of derivatives ofrational and radical functions using the definition in section 11. Review the algebra rules forexponents, and be very comfortable with the facts x1/n = n

px and x�n = 1

x

n . First rewrite fusing these rules.

pt� 1

3pt

= t1/2 � 1t

1/3

= t1/2 � t�1/3.

28

Then

f

0(t) =d

dt

�t

1/2 � t�1/3�

(We rewrote f .)

= (1/2)t�1/2 � (�1/3)t�4/3 (Bring the powers down and subtract 1 per the Power Rule.)

=1

2· 1t

1/2

+1

3· 1t

4/3

(We use algebra rules for exponents.)

=1

2pt

+1

3 3pt

4

.

3. y =5

x

2

.

Rewrite as y = 5x�2. Thendy

dx

= 5(�2)x�3 = �10x

3

.

4. f(x) =x

2x+ 1Use the Quotient Rule. We get

f

0(x) =

⇥d

dx

(x)⇤(2x+ 1)�

⇥d

dx

(2x+ 1)⇤(x)

(2x+ 1)2

=(1)(2x+ 1)� (2)(x)

(2x+ 1)2

=1

(2x+ 1)2.

5. s =pt(t3 + 1).

We use the Product Rule. I only show the first step and the answer here, so work out thealgebra on your own! (Ask for help if you get stuck.) The derivative is

ds

dt

= (1/2)t�1/2(t3 + 1) + t1/2(3t2 + 0) =t

3 + 1

2t1/2+ 3t5/2.

Personally I don’t care whether you write the final answer with p symbols or not. Note thatyou could distribute the

pt to simplify s before di↵erentiating, which obviates the need for

the Product Rule, and gives you the same answer. Wow!

13 Rates of Change

If you only remember one thing from this class, it should be that “the derivative is the slope of thetangent line,” or “the derivative is the instantaneous rate of change.” In fact, they are two di↵erentways to say the same thing.

29

For a function s(t) of time, the average rate of change on the interval a t b is

�s

�t=

s(b)� s(a)b� a .

The instantaneous rate of change of s at time t is s0(t).

The average rate of change is exactly the same as the slope of the line through (x, y) = (a, s(a))and (b, s(b)). Note how all the notation is starting to fit together (thanks to Leibniz), because

s

0(t) =ds

dt

= lim�t!0

�s

�t.

A derivative is a limit of an average rate of change. It makes sense that if you calculate average rateof change on an infinitesimally small interval (�t! 0) you get instantaneous change. That’s prettymuch why calculus was discovered. It allows the concepts of velocity and acceleration from physicsto be put in a nice mathematical language, and the math teaches us new things about physics.Calculus applies to every field of modern science however, so we talk about “rates of change” ratherthan just velocity. Most of applied math involves di↵erential equations, which are equations thatrelate quantities and their rates of change (a.k.a. derivatives). Chemical reaction rates, populationgrowth, and marginal profit are examples from chemistry, biology, and finance that are commonlystudied and explained with calculus. I would love to start working through some major applicationsright now, but most of the ones I can think of involve understanding the derivative of the exponentialfunction y = ex. Such courses would be “early transcendentals” courses, which are not this course.We have good reasons to do it our way, but life science and economics majors might consider specialcalculus courses for their interests (Mānoa has MATH 215 for bio majors and 203 for econ). Ourcourse is geared toward covering everyone’s needs, so we can’t skip over the physics and engineeringmajors, who need the most technical detail. The bio folks who mainly need the exponential functionwill be appeased finally in calc 2 at KCC.

Example: Galileo throws rocks o↵ of the leaning tower of Pisa and measures their fall over time.If s(t) is the height in feet of a rock after t seconds, he figured out that

s(t) = �16t2 + v0

t+ s0

.

This is the equation of a free-falling object near Earth’s surface. The velocity v = s0 is the derivativeof position (or displacement) and a = v0 = s00 is the acceleration. Note that v

0

= v(0), a constantrepresenting the initial velocity, and s

0

= s(0) is the initial position. We can use the rules fromsection 3.2 to easily find

v(t) = �32t+ v0

and a(t) = �32. The acceleration of Galileo’s rocks due to gravity is constant. Newton usedcalculus to explain his Laws of Motion and Law of Universal Gravitation, which was one of thegreatest scientific achievements of humankind.

Spoiler alert: Bio folks, what you will need the most in your field is solutions to di↵erentialequations like y0 = ky, for some constant k, which says the rate of change of y is proportional to its

30

size. If y is the number of organisms in a population, then this is a reasonable model for unrestrictedpopulation growth (no predators, disease, or lack of food). The solution to this di↵erential equationis y = ekt, where t is time and e is the transcental number e ⇡ 2.7. In Calc 2 you will learn thaty = ex is its own derivative (woah!) which is what gives exponential growth in this situation. Throwin predators or disease and the di↵erential equation changes, and so does the population function.We learn calculus in hopes to solve any di↵erential equation that arises in the science that we love,allowing us to analyze data and give a sound logical basis for our hypotheses.

14 Trigonometric Derivatives

In this section we harvest the fruit from the seeds that were planted in section 8. Mixing trigonom-etry and calculus can actually be quite nice, and the nicest fact is that d

dx

sin x = cosx. Here is aproof of the fact, using the definition of the derivative:

d

dx

sin x = limh!0

sin(x+ h)� sin xh

(1)

= limh!0

sin x cosh+ sinh cos x� sin xh

(2)

= limh!0

sin x cosh� sin x+ sinh cos xh

(3)

= limh!0

✓sin x cosh� sin x

h

+sinh cos x

h

◆(4)

= limh!0

sin x cosh� sin xh

+ limh!0

sinh cos x

h

(5)

= limh!0

sin x(cosh� 1)h

+ limh!0

sinh

h

(cos x) (6)

= (sin x) limh!0

(cosh� 1)h

+ (cosx) limh!0

sinh

h

(7)

= (sin x)(0) + (cos x)(1) (8)

= cosx. (9)

Line (1) is the definition of the derivative, and line (2) comes from the trig identity for summedangles. Lines (3)-(4) come from simple algebra. Line (5) comes from the sum law for limits. Line(6) is simple factoring. Line (7) is the constant multiple rule for limits; note that h is the variablein the limit so x is constant. Line (8) is the trig limits from section 8.

We just gave the analytic proof that ddx

sin x = cosx. The geometry behind it is beautiful, sincethe slopes of a sine wave make another sine wave (a cosine graph is just a shift of a sine graph).

31

slopes of f

⇡

1

y-values of f 0

⇡

1

Although the x and y axis are not quite to scale in the diagram above (which a↵ects slope), itlooks indeed like f 0(x) = cos x is the derivative of f(x) = sin x. From this, you can understand thatd

dx

cos x = � sin x. Check that the fourth derivative of sinx is sin x. Nice, right? This repetitive(periodic) behavior of trig functions under di↵erentiation makes them useful in solving di↵erentialequations, so even bio folks like to know that d

dx

sin x = cosx. In fact, with Euler’s formulae

i✓ = cos ✓ + i sin ✓, we see a connection between trig and the most important function in biology,f(x) = ex (actually the most important function in pretty much everything). The other trigfunctions’ derivatives are easy to derive from the sine and cosine. For example,

d

dx

tan x =d

dx

sin x

cos x(10)

=

�d

dx

sin x�cos x�

�d

dx

cos x�sin x

cos2 x(11)

=(cos x) cosx� (� sin x) sin x

cos2 x(12)

=cos2 x+ sin2 x

cos2 x(13)

=1

cos2 x= sec2 x. (14)

Line (11) is the quotient rule, and (14) comes from the Pythagorean trig identity. You shouldderive the rest of these for practice:

d

dx

sin x = cosxd

dx

cos x = � sin xd

dx

tan x = sec2 xd

dx

cot x = � csc2 xd

dx

sec x = secx tan xd

dx

csc x = � csc x cot x

32

15 The Chain Rule

The Chain Rule is listed in section 12, but here we will go into more detail and examples. It tends tobe something students don’t understand, so there are lots of cute applications people have thoughtup. Gears of di↵erent sizes turning each other, Americans driving cars in Canada (so they have toconvert to km) and party balloons all involve situations where the chain rule can be experiencedin real life. However, most students are still wrapping their heads around derivatives as rates ofchange, so most people learn the computational aspect of the chain rule before they understand theapplications.

Review function composition from precalculus. When one function is composed with another,it gets “plugged into” the other. I.e., (f � g)(x) = f(g(x)). The little � indicates composition, andwe say “f composed with g” or “f of g of x”.

The Chain RuleIf y = f(u) and u = g(x) are di↵erentiable functions, then

d

dx

f(g(x)) = f 0(g(x)) · g0(x) and dydx

=dy

du

· dudx

.

Leibniz’s di↵erential notation works extremely well upon consideration of the chain rule (it’ssort of like the dus cancel; wow!). In practice, students might get stumped by the notation, butthere is a decent way to describe the chain rule in our vernacular: Take the derivative of the outside,leaving the inside alone, then multiply by the derivative of the inside.

Examples.d

dx

(x2)5 = 5(x2)4 · (2x) = 10x9.

d

dx

sin(3x) = cos(3x) · 3 = 3 cos 3x.

d

dx

qtan4(x7) =

1

2

�tan4(x7)

��1/2 · (4 tan3(x7)) · (sec2(x7)) · 7x6

= 14x6 tan(x7) sec2(x7).

The first example above is evidence that the Chain Rule works. Note that (x2)5 = x10, so ifwe simplify first we can use the Power Rule. When there are parentheses in a function’s formula,the chain rule comes into play. Students who start using the Chain Rule sometimes don’t know“when to stop” di↵erentiating. The number of parentheses indicates the number of factors in theresult. Note that in the last example, there are 3 sets of parentheses because two are hidden:

(tan4(x7))1/2 =⇣(tan(x7))4

⌘1/2

. The number of pairs of parentheses in the function is the number

of multiplication dots in the derivative (before simplifying) per the Chain Rule. We could have

simplified the last example before di↵erentiating, since⇣(tan(x7))4

⌘1/2

= tan2(x7).

33

Just in case you want a deeper understanding of the chain rule, the following is an analyticargument for it. Assume g is di↵erentiable (hence continuous) but not constant near a, and use the(alternate) definition of the derivative.

d

da

f(g(a)) = limx!a

f(g(x))� f(g(a))x� a

= limx!a

f(g(x))� f(g(a))x� a ·

g(x)� g(a)g(x)� g(a)

= limg(x)!g(a)

f(g(x))� f(g(a))g(x)� g(a) · limx!a

g(x)� g(a)x� a

= f 0(g(a)) · g0(a),

provided the limits exist.

16 Implicit Di↵erentiation

In precalculus we learn that a set of points is a relation, and a function is a relation that passesthe vertical line test (v.l.t.). For example, the points (x, y) that satisfy x2 + y2 = 1 are a relationthat fails the v.l.t.. It turns out that there is a nice way to find slopes dy

dx

to this relation anyway.We have only defined di↵erentiation for functions, so to extend our definition to relations that arenot functions, we carefully examine small neighborhoods on a graph, where (hopefully) the relationpasses the v.l.t. and defines a function implicitly. I won’t go into the technicalities of the vocabularyand worry about implicit vs. explicit here. It su�ces to say you need the Chain Rule, and thingswork out neatly.

Implicit Di↵erentiation

Di↵erentiate both sides of the equation with respect to x, treating y as a di↵erentiable

function of x. From the Chain Rule, a factor ofdy

dx

appears in every expression with y in it.

Example: x2 + y2 = 1. The graph of this relation is the unit circle. We will find its slope at anypoint (x, y) using implicit di↵erentiation.

34

x

2 + y2 = 1 (1)

d

dx

(x2 + y2) =d

dx

1 (2)

d

dx

x

2 +d

dx

y

2 = 0 (3)

2x+ 2y · dydx

= 0 (4)

2ydy

dx

= �2x (5)dy

dx

=�2x2y

(6)

dy

dx

=�xy

(7)

The most di�cult step for most students to grasp is (4). Since y2 is a composite function, we

need the Chain Rule. However, we don’t need it for x2, although technically writing 2xdx

dx

as the

first term in (4) is correct (but silly because dxdx

= 1). In y2, the outside function is the squaring

function, and the inside function is y. The cool outcome is thatdy

dx

=�xy

gives us the slope at any

point on the unit circle.

x

slopes dydx

= �xy

y

⇣p2

2

,

p2

2

⌘

(�1, 0)

The slope at (x, y) =⇣p

2

2

,

p2

2

⌘is

dy

dx

=�xy

= �p2

2p2

2

= �1, which is obvious in the picture.

Wow! And at (�1, 0) the slope dydx

= �xy

=1

0is undefined, which matches the picture since there

is a vertical tangent line there.

35

Example: xy = 1. This relation is actually a function because its graph passes the v.l.t.. If you

solve for y you see y = f(x) =1

x

. Therefore, by the Power Rule, y0 =�1x

2

. However, we can also

use implicit di↵erentiation. We use the product rule in the following:

xy = 1

d

dx

(xy) =d

dx

1✓

d

dx

x

◆(y) +

✓d

dx

y

◆(x) = 0 (Product Rule)

(1) (y) + (y0)(x) = 0

y + y0x = 0

y

0 =�yx

In fact this is the same answer we got by solving for y first, because

y

0 =�yx

=��1

x

�

x

= �✓1

x

◆· 1x

=�1x

2

.

Wow!

17 Related Rates

Check out this cistern for catching drinkable rain water (from the College of Tropical Agricultureand Human Resources at Mānoa).

If two quantities are related, such as volume and height of a cylinder, then the rates of changeof the quantities are related. An equation determines the relation, and di↵erentiating gives a

36

relationship between the derivatives, a.k.a. rates of change. Once again, if there is only one thingyou remember from this class, it should be that derivatives are instantaneous rates of change.

Related Rates

Di↵erentiate both sides of the relation with respect to t (time), using the Chain Rule for

composite functions. Understand that if A represents a quantity, thendA

dt

represents the rate

of change of that quantity.

Getting an equation with derivatives in it allows us to solve interesting problems. (Oh boy!Word problems! Now we can use the math in real life!;)) These 6 steps are the same steps I giveall my students in all my classes, and I give partial credit based on following the steps. Of course,step 4 is special for related rates problems.

1. Read the problem at least 3 times.

2. Label the variables. (Write down what’s what. Easy partial credit!)

3. Write an equation relating the variables.

4. Solve: Di↵erentiate both sides of the relation with respect to time t, using the Chain Rule for

composite functions. Understand that if A represents a quantity, thendA

dt

represents the rate

of change of that quantity. Plug in the given information and solve for the unknown.

5. State your answer in a sentence.

6. Check for reasonableness.

Example. The cistern in the photo is 3m in diameter. Rain is falling at 3cm/hr straight into thecistern (not how they really work, more below). How fast is the volume of water in the cisternincreasing?

Step by step solution.

1. Read it three times. The quantities mentioned are diameter and volume. Diameter is constant,so we don’t have to label it a variable. However, the height of water is changing at 3cm/hr,so height should be a variable. The question we need to answer is about the rate of change ofvolume.

2. Let h be the height of the water in the cistern, and let V be the volume of water in the cistern.(Write this on a test for easy partial credit.) Since rain is falling at 3cm/hr, the height h is

increasing 0.03m/hr, which meansdh

dt

= 0.03. Since we want to find the rate of change of

volume, the unknown isdV

dt

.

37

3. We can tell from the picture that the cistern is a right circular cylinder (or close, but ona test it would be stated explicitly) so from our knowledge of geometry V = ⇡r2h wherer = 3/2 = 1.5, giving r2 = 2.25 (square meters). Hence,

V = 2.25⇡h.

4. Di↵erentiate w.r.t. t:

V = 2.25⇡h (8)

d

dt

V =d

dt

(2.25⇡h) (9)

dV

dt

= 2.25⇡dh

dt

. (10)

All we used to get (3) is the Constant Multiple Rule for derivatives. This example is so simplethat we do not need the Chain Rule. However, if the equation were more complicated, likeif h was raised to a power other than 1, eg., we would need the Chain Rule. Now plug indh

dt

= 0.03 and solve fordV

dt

.

dV

dt

= 2.25⇡(0.03) ⇡ 0.212.

5. The volume is increasing at about 0.212 m3/hr. PAU. (That’s the answer.)

6. Let’s think about it. If the height of the cistern (max height of the water) is 1m (a reasonableestimate), the total volume of the container is ⇡(1.5)2(1) ⇡ 7 m3. It is going to take a coupledays to fill up (if the rain keeps coming), which is reasonable.

The truth is, we could probably solve this problem without calculus, since the relationshipbetween volume and height is linear (a.k.a. direct variation). Calculus will be needed for morecomplicated equations though, like when the Chain Rule comes into play. Also, cisterns aren’tjust buckets open to the rain, they collect water through pipes from the roof of a building.So another related rate problem is to compare the area of the roof to the change in volumein the tank. This one is just as easy mathematically, so we will do a di↵erent one instead.

Example: A 16m ladder is leaning against a building, and being pulled up. A worker ispulling the top of the ladder up at 0.5m/sec. How fast is the bottom of the ladder movingalong the ground at the moment when its distance from the building is 8m?

38

buildingladder

0.5m/s

dr

dt The picture is a right triangle with hypotenuse 16, bot-tom leg r, and let’s call the side leg x. The Pythagorean Theorem gives r2 + x2 = 162.Then

2rdr

dt

+ 2xdx

dt

= 0.

Since the worker is pulling up, dxdt

= 0.5 m/sec. We are given r = 8 so we need to find x.

Using the Pythagorean Formula, x2 = 162 � 82, so x =p192 = 8

p3. Plugging it all in,

2(8)dr

dt

+ 2(0.5)(8p3) = 0

dr

dt

=�8p3

16dr

dt

= �p3

2

The rate drdt

is negative because the distance r is decreasing. The ladder is sliding along the

ground atp3

2

⇡ 0.866 m/sec.

18 Linear Approximation and Di↵erentials

So far in this chapter, each section has depended on or been related to the previous one. Thissection does not depend on the previous section, so it will seem a bit random (miscellaneous).

Linear approximation is simply approximating functions by their tangent lines.

The linear approximation of f(x) at x = a is

L(x) = f 0(a)(x� a) + f(a).

The function L is literally the tangent line to f(x) at a. We found tangent lines earlier (as farback as section 2) with point slope form y � y

0

= m(x� x0

), and in the framed equation above weare just rewriting this while setting x

0

= a (so y0

= f(x0

) = f(a)), y = L(x), and m = f 0(a).

39

Example: The linear approximation to f(x) = 3px at x = 1 is

L(x) =1

3(x� 1) + 1.

All we did was plug in a = 1, f 0(a) = 13

(a)�2/3 = 13

(1)�2/3 = 13

, and f(a) = 3p1 = 1. Suppose

you were stranded on a desert island without a calculator (the kind of thing that happens all thetime on television), and you had to figure out the cube root of 1006 (for some engineering probleminvolving opening a coconut or something). Suppose also that you have just learned about linearapproximation. It’s not that di�cult to do the following calculation in the sand (or even your head):

3p1006 = 10 3

p1.006 = 10f(1.006) ⇡ 10L(1.006) = 10

1

3(1.006� 1) + 1

�= 10[0.002 + 1] = 10.02.

Watching you on TV, the audience is wowed when they check with their calculators, 3p1006 =

10.0199601328.... Your mental calculation is accurate within 0.00004, and the coconut slingshotworks amazingly well.

All jokes aside, linear approximation is a handy tool. It works best when you are approximatingnear the base point a. The idea is that the tangent line is close to the function there, as we see inthis graph of y = 3

px and its tangent line at a = 1.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

The change in f from its base point to its value at x is close to the change in the tangent line,which is the slope times the change in x. Look back at the definition of the derivative, but let’s put�x as the change in x, instead of h.

f

0(x) = lim�x!0

f(x+�x)� f(x)�x

.

Then since �y = f(x+�x)� f(x), we can write

f

0(x) = lim�x!0

�y

�x=

dy

dx

.

This explains Leibniz’s notation f 0(x) = dydx

. For certain purposes, it helps to “separate the

fraction,” but this will confuse students because a derivativedy

dx

is not a fraction in the usual sense.

40

Di↵erentialsIf y = f(x) is a di↵erentiable function and dx is a real number, the di↵erential, dy, is defined as

dy = f 0(x)dx.

The idea is that dy ⇡ �y. Note that dy is NOT the derivative y0. (Quick vocab review:“derivative” 6= “di↵erential”, but “the derivative of f exists” = “f is di↵erentiable”.) The confusingthing for students is that we are talking about infinitesimally small things here, so limits are behindthe scenes. We have to use a real number as dx when we define the di↵erential dy, but we areintending to let dx approach 0 when we use di↵erentials. This stu↵ will come up again in Chapter5, but for the Chapter 3 test all I ask is that you know how to do easy problems like this:

Example: For y = f(x) = x4 + x and a real number dx, find the di↵erential dy.

Di↵erentiate, writingdy

dx

= 4x3 + 1, and the answer is then

dy = (4x3 + 1)dx.

19 The Extreme Value Theorem

Chapter 4, this chapter, is a collection of powerful tools that come from advanced use of derivatives.We already learned about them as slopes or rates of change. Now, we apply these ideas to help usgraph functions and find extrema. The word extremum (the plural is extrema) means maximum orminimum. We will talk about maxima and minima a lot in this chapter. I will probably abbreviatewith “max” or “min” often.

The Extreme Value TheoremIf f is continuous on [a, b], then f has a maximum value and a minimum value on [a, b].

When we say “value” we mean y-value. The Extreme Value Theorem is important to thetheoretical framework of calculus. It is interesting to see examples of functions that do not havemaxima or minima, and note that none of them are continuous.

It is imperative to understand the meaning of the open dot. The function values get infinitelyclose to the y-value of the dot, but never reach it. Hence, maxes or mins can fail to be achieved.Asymptotes can also ruin a function’s chance to have extreme values, but functions are nevercontinuous at their vertical asymptotes.

Anyway, continuous functions on closed intervals always have a max and a min, and we wantto find them. Think about it; our life goals are to MAXIMIZE AWESOMENESS, MINIMIZELAMENESS. That is what this chapter is all about. The key idea is that extrema occur at eithercorners or horizontal tangents. We call these critical points.

41

x

y

a

b

No max or min.

x

y

a

b

Two maxima, no mininmum.

The Critical Point TheoremA critical point is an x-value x = c in the domain of f such that f 0(c) = 0 or f 0(c) is

undefined. The function f(x) can only have extreme values at critical points or endpoints.

We have to include the endpoints x = a and x = b when we look at functions on [a, b].

Example. Find the extreme values of f(x) = x2 � x on [�1, 3].The candidates for locations of a max or min are the endpoints �1 and 3, and the critical points.

Since f 0(x) = 2x � 1, the only critical point is when 2x � 1 = 0, so x = 12

. Plug each candidatex-value into the function y = f(x), and the y-values will determine the max and min.

f(�1) = (�1)2 � (�1) = 2f

�1

2

�=�1

2

�2 � 1

2

= 14

� 12

= �14

f(3) = 9� 3 = 6Answer: The max is 6 and the min is �1

4

.Crucial to understanding the Critical Point Theorem is the visual

42

x

y

horizontal tangent line at critical point, min

which shows the min at the vertex of this parabola where the tangent line is horizontal. In thischapter, I will frequently say “set the derivative to 0,” and this is why.

More on extrema:We did not give the technical definition of a maximum or minimum above. The idea is usually

quite intuitive, but there are some situations in which we must know the technical definition, asfollows:

For a function f(x), the point (a, f(a)) is an absolute maximum if f(a) � f(x) for everyx in the domain of f . The point (a, f(a)) is a local maximum if f(a) � f(x) for all x in someneighborhood (like a small open interval) of a.

For a function f(x), the point (a, f(a)) is an absolute minimum if f(a) f(x) for everyx in the domain of f . The point (a, f(a)) is a local minimum if f(a) f(x) for all x in someneighborhood of a.

First, note the inequality is not strict, i.e., if f(a) = f(x) for all x, then f has a maximumat a. So for a constant function, every single point on the graph is both a max and min! As forthe di↵erence between “absolute” and “local” (a.k.a. “relative”), a picture (below) is helpful. Theword “neighborhood” is actually a profound word in mathematics, and we will not get into it inthis class (a starting point would be section 2.3). Some textbooks di↵er slightly on the definitionof a local extremum, especially regarding whether endpoints of a function’s domain can be thelocation of local extrema. In math, you can define things however you want as long as you areconsistent throughout, but I will avoid the local-max-at-endpoint issue here. The important thingis to understand the following picture.

43

x

y

min

local max

local min

no absolute max

Understand that an absolute max can just be called a max, but if a point is a local max, andnot absolute, the word local must be emphasized. The word relative is also used, interchangeablywith local.

20 The Mean Value Theorem

The Mean Value Theorem (MVT) is a big part of the theoretical framework of calculus. Manyproofs of subsequent theorems rely on the Mean Value Theorem, including facts about increas-ing/decreasing functions and the Fundamental Theorem of Calculus. In advanced calculus whererigorous proofs are done (like MATH 331 at Mānoa), the MVT saves the day time and time again.At this level, beginning calculus, it is di�cult to appreciate the MVT, but I ask that you try tounderstand the picture. If nothing else yet, the MVT exercises are good for practicing the languageof calculus, and solidify your understanding of other concepts as well.

The Mean Value TheoremIf f is continuous on [a, b] and di↵erentiable on the interval (a, b), then there exists a number

c in (a, b) such that

f

0(c) =f(b)� f(a)

b� a .

In less formal terms, the MVT says “there is a place where the tangent line is parallel to the

secant line,” because f 0(c) is the tangent slope andf(b)� f(a)

b� a is the secant slope. Here’s thepicture:

44

x

y

a

b

c

f

tangent

secant

Example. Explain why the Mean Value Theorem applies to f(x) =px on [0,9], and find the

number c guaranteed by the Theorem.SOLUTION: The function f(x) is continous on [0, 9] and di↵erentiable on (0, 9), so the Mean

Value Theorem applies. It says we can find c such that

f

0(c) =f(9)� f(0)

9� 0 =p9�p0

9� 0 =1

3.

Since f 0(x) = 12

x

�1/2, we solve

1

2x

�1/2 =1

3.

Multiply both sides by 2 to get

x

�1/2 =2

3,

and raise both sides to the �2 to get

x =

✓2

3

◆�2=

✓3

2

◆2

=9

4.

So c = 94

.

21 Increasing and Decreasing Functions and the First Deriva-tive Test

In MATH 135 at KCC we learn how to tell when a function is increasing, decreasing or constant byits graph. The keys are to read left to right, and to only give intervals on the x-axis. For example,consider the graph of y = f(x) below.

45

x

y

(�2, 1)

This function f is increasing on (�1,�2], (�2, 0], decreasing on [2,1), and constant on [0, 2].Note that we don’t give any y-values when we talk about increasing and decreasing, and we do

not use [ signs. The formal definition of “increasing” is that f(x2

) > f(x1

) whenever x2

> x

1

.The graph above suddenly jumps down at x = �2, so it would be wrong to say that f increases on(�1, 0].

WRONG: f is increasing on (�1,�2] [ (�2, 0].CORRECT: f is increasing on (�1,�2], (�2, 0].The experienced student will know that a line with negative slope is decreasing, and a line with

positive slope is increasing. Hence, the relationship with the derivative f 0 is not surprising:

If f 0(x) > 0 on the interval (a, b), then f(x) is increasing on (a, b).If f 0(x) < 0 on (a, b), then f(x) is decreasing on (a, b).If f 0(x) = 0 on (a, b), then f(x) is constant on (a, b).

Like I say, the framed result above is not surprising. A rigorous proof, however, is one of thefirst places where the Mean Value Theorem saves the day. Here is what that looks like:

Theorem. If f 0(x) > 0 on the interval (a, b), then f is increasing on (a, b).

Proof. Suppose f 0(c) > 0 at every point c in between a and b. Given x2

> x

1

in the interval, wecan apply the Mean Value Theorem to find c such that

f

0(c) =f(x

2

)� f(x1

)

x

2

� x1

.

Hencef(x

2

)� f(x1

)

x

2

� x1

> 0.

46

Multiplying both sides of the inequality by x2

� x1

gives

f(x2

)� f(x1

) > 0,

which means f(x2

) > f(x1

). Hence f is increasing.

Now, recall that the major goal of this chapter (and life) is to MAXIMIZE AWESOMENESS,MINIMIZE LAMENESS. How do the ideas of increasing/decreasing functions do this? The answeris the First Derivative Test.

The First Derivative TestLet x = c be a critical point of f .If f 0(x) changes from + to � at c, then f has a local max at c.If f 0(x) changes from � to + at c, then f has a local min at c.

Example: Find the intervals of inc/dec and local extrema of f(x) = x3 � 3x.Apply the test-point method or a sign chart (precalculus skills) to find the sign of the derivative

f

0(x) = 3x2 � 3 = 3(x� 1)(x+ 1).

�1 1

f(x) inc incdec

f

0(x) + + + +++���

Thus, f(x) is increasing on (�1,�1), (1,1), and decreasing on (�1, 1). (You could use somebrackets if you know how; di↵erent teachers may specify the forms of intervals they want you towrite.)

There is a local max at x = �1 and a local min at x = 1, by the First Derivative Test. Personally,I take points o↵ if you omit the “at” because the max value is a y-value, not �1. To be safe, youcould plug the values into the original function, and write “the point (�1, 2) is a local max of f ,and (1,�2) is a local min.”

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

47

22 Concavity

Look at this picture of a concave up function, where we estimate slopes from left to right.

slope ⇡ �3 �13

0 13

3 increasing

This is the visual idea behind concavity. If the slopes of f are increasing, then f 0 is increasing.Since f 0 is increasing, its derivative is positive. The derivative of f 0 is the second derivative of f .

If f 00(x) > 0 on some interval (a, b), then f(x) is concave up on (a, b).If f 00(x) < 0 on (a, b), then f(x) is concave down on (a, b).A number x = c where f 00(c) = 0 and f 00 changes sign is called an inflection point.

Note: Technically f 00 may be undefined at an inflection point, but f 0 has to be defined.Students will learn to understand that the concavity of f is not related to whether f is increasing

or decreasing. The following table shows four types of functions. The abbreviations ccu and ccdmean concave up and concave down, respectively.

inc

dec

ccu ccd

48

In precalculus, you learn tools and techniques for graphing, like how to plot points, find interceptsand asymptotes, and determine end behavior. With the ideas of increasing/decreasing and concaveup/down, calculus gives us new tools to add to our ability to graph. It is hard to overestimate howimportant graphing is in science and technology. Obviously the human brain processes picturesdi↵erently than equations, but graphing connects them. A picture is worth a bazillion words.

23 Optimization

This is a section that students tend to fear, because it consists of word problems. We are goingto make it easy and friendly. The examples will all follow the same pattern, and the key step willalways be the same: “set the derivative to 0.” Rigorous mathematics texts will emphasize the otherways that maxima and minima arise (at corners where f 0 is undefined, at endpoints of the domain,and at points other than where you find f 0 is 0), but for our purposes, in the context of the exampleshere, setting the derivative to 0 is enough. The magic of optimization is prettiest when f 0 is 0, andthat is also the most common way optimization is used. So sit back, relax, and get ready to setsome derivatives to 0.

OptimizationPlug the constraint into the objective, to get a function of one variable. Find the critical

point by setting the derivative to 0. Answer the question.

The examples will all have an objective function and a constraint equation. The objective is thething that needs to be maximized or minimized (optimized). The constraint relates two variables.

Example: Find the maximum product of two numbers whose sum is 110.

Let x and y be the two numbers. The objective is the product P = xy. The constraint isx+ y = 110. Plug the constraint x+ y = 110) y = 110� x into the objective

P = x(110� x) = 110x� x2.

Then set the derivative to 0.

P

0 = 110� 2x = 0) 110 = 2x) x = 55.

The product is maximized when x = 55 and y = 110 � x = 110 � 55 = 55. The maximumproduct is (55)(55) = 3025.

Example: You want to enclose your cows next to a cli↵. Maximize the area inside a 3-sidedrectangular fence (see picture) with 600 total meters of fence.

49

x

ycows in here

overhead view

The constraint is 2x+ y = 600, which is the perimeter of 3 sides of the rectangle. The objectiveis the area A = xy. Plug the constraint 2x+ y = 600) y = 600� 2x into the objective

A = xy = x(600� 2x) = 600x� 2x2.

Then set the derivative to 0.

A

0 = 600� 4x = 0) 600 = 4x) 150 = x.

The area is maximized when x = 150 and y = 600�2(150) = 300. The max area is (150)(300) =45000 square meters.

Example: You are selling x bento lunches at price p. It costs you $5 to make one. By supplyand demand x = 20 � p (if you raise the price, you sell less). What price should you charge tomaximize the profit P = xp� 5x?

This is a standard business math problem. The objective is profit P = xp � 5x, which comesfrom revenue xp minus cost 5x. The constraint is x = 20�p. Plug the constraint into the objective

P = (20� p)p� 5(20� p) = 20p� p2 � 100 + 5p = �p2 + 25p� 100.

Then set the derivative to 0.

P

0 = �2p+ 25 = 0) 25 = 2p) 12.5 = p.

The price you should charge is $12.50.

Notes: One thing that makes these examples easy is that they all have only one critical point.Sometimes there are two or more, and then you have to figure out which one(s) answer the question.The gnarly optimization problems just take cleverness to figure out what the constraint and objectiveare, and which critical point answers the question (if it even has an answer).

24 Newton’s Method

Newton’s Method is a way to approximate roots of functions. Recall that a root of f(x) is a numbersuch that f(r) = 0. In precalculus you learn the quadratic formula to find roots of quadratics

50

and synthetic division to search for roots of high degree polynomials, but in general it is not easyto find roots. Nowadays, we use calculators and computers. In Newton’s time, any method thatmade calculations easier was a big deal, and today we still need people who know how to programthe calculators we use. Newton’s method boils down to a straightforward formula, but a calculusstudent will benefit their overall understanding if they can derive the formula themselves, ratherthan memorize it. It is simply a tangent line approximation. Since a root r gives an x-intercept ofthe graph of f(x), Newton’s Method is: find the x-intercept of a tangent line to f near its root. Youstart by picking a number x

0

that is (already) an approximation of r. Then you find the tangentline at x

0

, and its x-intercept is x1

.f

r x

0

(x0

, f(x0

))

x

1

The derivation of the formula for x1

is something a good calculus student will know how to do.You start with the point-slope form of the tangent line at x

0

, which is

y � f(x0

) = f 0(x0

)(x� x0

).

Then you find the x-intercept by setting y = 0 and solving for x:

0� f(x0

) = f 0(x0

)(x� x0

)

�f(x0

) = f 0(x0

)x� f 0(x0

)x0

f

0(x0

)x0

� f(x0

) = f 0(x0

)x

f

0(x0

)x0

� f(x0

)

f

0(x0

)= x

x

0

� f(x0)f

0(x0

)= x = x

1

.

Then you iterate. Iteration is a big idea in mathematics. It means “repeat the formula,” or “run itthrough the function again,” in layman’s terms. You take the x

1

value and find a new tangent lineat x

1

whose x-intercept x2

is an even better (hopefully) approximation of r:

x

1

� f(x1)f

0(x1

)= x

2

.

The process can be repeated until your approximation is as close as you need. Most studen