Introduction to Calculusweb2.hunterspt-h.schools.nsw.edu.au/studentshared/MATHEMATICS… ·...

TERMINOLOGY

8 Introduction to Calculus

Composite function: A function of a function. One function, f (x), is a composite of one function to another function, for example g(x)

Continuity: Describing a line or curve that is unbroken over its domain

Continuous function: A function is continuous over an interval if it has no break in its graph. For every x value on the graph the limit exists and equals the function value

Derivative at a point: This is the gradient of a curve at a particular point

Derivative function: The gradient function of a curve obtained through differentiation

Differentiable function: A function which is continuous and where the gradient exists at all points on the function

Differentiation: The process of fi nding the gradient of a tangent to a curve which is called the derivative

Differentiation from fi rst principles: The process of fi nding the gradient of a tangent to a curve by fi nding the gradient of the secant between two points and fi nding the limit as the secant becomes a tangent

Gradient of a secant: The gradient (slope) of the line between two points that lies close together on a function

Gradient of a tangent: The gradient (slope) of a line that is a tangent to the curve at a point on a function. It is the derivative of the function

Rate of change: The rate at which the dependent variable changes as the independent variable changes

ch8.indd 438 6/29/09 12:46:04 PM

439Chapter 8 Introduction to Calculus

INTRODUCTION

CALCULUS IS A VERY IMPORTANT part of mathematics and involves the measurement of change. It can be applied to many areas such as science, economics, engineering, astronomy, sociology and medicine. We also see articles in newspapers every day that involve change: the spread of infectious diseases, population growth, infl ation, unemployment, fi lling of our water reservoirs.

For example, this graph shows the change in crude oil production in Iran over the years. Notice that while the graph shows that production is increasing over recent years, the rate at which it is being produced seems to be slowing down. Calculus is used to look at these trends and predict what will happen in the future.

There are two main branches of calculus. Differentiation is used to calculate the rate at which two variables change in relation to one another. Anti-differentiation, or integration, is the inverse of differentiation and uses information about rates of change to go back and examine the original variables. Integration can also be used to fi nd areas of curved objects.

DID YOU KNOW?

‘Calculus’ comes from the Latin meaning pebble or small stone. In ancient civilisations, stones were used for counting. However, the mathematics practised by these early people was quite sophisticated. For example, the ancient Greeks used sums of rectangles to estimate areas of curved fi gures.

However, it wasn’t until the 17 th century that there was a breakthrough in calculus when scientists were searching for ways of measuring motion of objects such as planets, pendulums and projectiles.

Isaac Newton, an Englishman, discovered the main principles of calculus when he was 23 years old. At this time an epidemic of bubonic plague closed Cambridge University where he was studying, so many of his discoveries were made at home.

He fi rst wrote about his calculus methods, which he called fl uxions, in 1671, but his Method of fl uxions was not published until 1704.

Gottfried Leibniz (1646–1716), in Germany, was also studying the same methods and there was intense rivalry between the two countries over who was fi rst!

Search the Internet for further details on these two famous mathematicians. You can fi nd out about the history of calculus and why it was necessary for mathematicians all those years ago to invent it.

7,000

Crude Oil Production (Mbbl/d)Iran

6,000

5,000

4,000

2,000

3,000

1,000

073

7475 77

76 7879 81 83 85 87 89

80 82 84 86 8891 93 95 97 99

90 92 94 96 9801 03 05 07

00 02 04 06

Tho

usan

d B

arre

ls p

er D

ay

January 1973–May 2007

You will learn about integration in the

HSC Course.

Isaac Newton

ch8.indd 439 8/1/09 10:50:59 PM

440 Maths In Focus Mathematics Extension 1 Preliminary Course

Gradient

Gradient of a straight line

The gradient of a straight line measures its slope. You studied gradient in the last chapter.

runrisem =

Class Discussion

Remember that an increasing line has a positive gradient and a decreasing line has a negative gradient.

positive negative

Notice also that a horizontal line has zero gradient.Can you see why?

Can you fi nd the gradient of a vertical line? Why?

Gradient plays an important part, not just in mathematics, but in many areas including science, business, medicine and engineering. It is used everywhere we want to fi nd rates.

On a graph, the gradient measures the rate of change of the dependent variable with respect to the change in the independent variable.

In this chapter you will learn about differentiation, which measures the rate of change of one variable with respect to another.

ch8.indd 440 6/29/09 12:46:58 PM


EXAMPLES

1. The graph shows the average distance travelled by a car over time. Find the gradient and describe it as a rate.

Hours

km

400

5t

d

Solution

The line is increasing so it will have a positive gradient.

runrisem

5400

180

80

=

=

=

=

This means that the car is travelling at the rate of 80 km/hour.

2. The graph shows the number of cases of fl u reported in a town over several weeks.

Weeks

Num

ber

ofca

ses

(100

s)

15

10t

N

Find the gradient and describe it as a rate.

CONTINUED

ch8.indd 441 7/10/09 11:59:24 AM


Solution

The line is decreasing so it will have a negative gradient. m

101500

1150

150

runrise

=

= -

= -

= -

This means that the rate is 150- cases/week, or the number of cases reported is decreasing by 150 cases/week.

When fi nding the gradient of a straight line in the number plane, we think of a change in y values as x changes. The gradients in the examples above show rates of change .

However, in most examples in real life, the rate of change will vary. For example, a car would speed up and slow down depending on where it is in relation to other cars, traffi c light signals and changing speed limits.

Class Discussion

The two graphs show the distance that a bicycle travels over time. One is a straight line and the other is a curve.

d

t

Hours

km

20

15

10

5

4321

t

Hours

km

20

15

10

5

4321

d

Is the average speed of the bicycle the same in both cases? What is different about the speed in the two graphs?

How could you measure the speed in the second graph at any one time? Does it change? If so, how does it change?

Gradient of a curve

ch8.indd 442 6/29/09 12:51:52 PM


Here is a more general curve. What could you say about its gradient? How does it change along the curve?

y

x

Copy the graph and mark on it where the gradient is positive, negative and zero.

Using what we know about the gradient of a straight line, we can see where the gradient of a curve is positive, negative or zero by drawing tangents to the curve in different places around the curve.

0

+-

y

x

Notice that when the curve increases it has a positive gradient, when it decreases it has a negative gradient and when it turns around the gradient is zero.

Investigation

There are some excellent computer programs that will draw tangents to a curve and then sketch the gradient curve. One of these is Geometer Sketchpad.

Explore how to sketch gradient functions using this or a similar program as you look at the examples below.

ch8.indd 443 6/29/09 12:52:05 PM


EXAMPLES

Describe the gradient of each curve.

1.

Solution

Where the curve increases, the gradient is positive. Where it decreases, it is negative. Where it turns around, it has a zero gradient.

2.

Solution

ch8.indd 444 7/10/09 11:59:47 AM


There are computer programs that will

draw these tangents.

EXAMPLE

Make an accurate sketch of (a) y x2= on graph paper. Draw tangents to this curve at the points where (b)

3, 2, 1, 0, 1, 2x x x x x x= - = - = - = = = and 3x = . Find the gradient of each of these tangents. (c) Draw the graph of the gradients (the gradient function) on a (d)

number plane.

Solution

(a) and (b)

9

8

7

6

5

4

3

2

1

1 2 3-3 -2x

y

(c)

3, 6

2, 4

1, 2

0, 0

1, 2

2, 4

3, 6

x m

x m

x m

x m

x m

x m

x m

At

At

At

At

At

At

At

= - = -

= - = -

= - = -

= =

= =

= =

= =

(d)

Since we have a formula for fi nding the gradient of a straight line, we fi nd the gradient of a curve by measuring the gradient of a tangent to the curve.

Use the ‘m’ values as the ‘y’ values on this

graph.

ch8.indd 445 7/10/09 12:00:15 PM


Drawing tangents to a curve is diffi cult. We can do a rough sketch of the gradient function of a curve without knowing the actual values of the gradients of the tangents.

To do this, notice in the example above that where m is positive, the gradient function is above the x -axis, where 0,m = the gradient function is on the x -axis and where m is negative, the gradient function is below the x -axis.

EXAMPLES

Sketch the gradient function of each curve.

1.

Solution

First we mark in where the gradient is positive, negative and zero.

Now on the gradient graph, place the points where 0m = on the x -axis. These are at ,x1 x2 and .x3

ch8.indd 446 6/29/09 12:52:40 PM


To the left of ,x1 the gradient is negative, so this part of the graph will be below the x -axis. Between x1 and ,x2 the gradient is positive, so the graph will be above the x -axis. Between x2 and ,x3 the gradient is negative, so the graph will be below the x -axis. To the right of ,x3 the gradient is positive, so this part of the graph will be above the x -axis.

2.

Solution

First mark in where the gradient is positive, negative and zero.

CONTINUED

ch8.indd 447 6/29/09 12:52:52 PM


8.1 Exercises

Sketch the gradient function for each graph.

The gradient is zero at x1 and .x2 These points will be on the x -axis. To the left of ,x1 the gradient is positive, so this part of the graph will be above the x -axis. Between x1 and ,x2 the gradient is negative, so the graph will be below the x -axis. To the right of ,x2 the gradient is positive, so this part of the graph will be above the x -axis.

1.

2.

3.

4.

5.

6.

ch8.indd 448 6/29/09 12:53:03 PM


7.

8.

9.

10.

Differentiation from First Principles

Seeing where the gradient of a curve is positive, negative or zero is a good fi rst step, but there are methods to fi nd a formula for the gradient of a tangent to a curve.

The process of fi nding the gradient of a tangent is called differentiation . The resulting function is called the derivative .

Differentiability

A function is called a differentiable function if the gradient of the tangent can be found.

There are some graphs that are not differentiable in places. Most functions are continuous , which means that they have a smooth

unbroken line or curve. However, some have a gap, or discontinuity, in the graph (e.g. hyperbola). This can be shown by an asymptote or a ‘hole’ in the graph. We cannot fi nd the gradient of a tangent to the curve at a point that doesn’t exist! So the function is not differentiable at the point of discontinuity.

a

y

x

This function is not differentiable at a since the curve is discontinuous at this point.

ch8.indd 449 7/10/09 12:00:30 PM


A function may be continuous but not smooth. It may have a sharp corner. Can you see why curves are not differentiable at the point where there is a corner?

A function ( )y f x= is differentiable at the point x a= if the derivative exists at that point. This can only happen if the function is continuous and smooth at .x a=

b

y

x

This function is not differentiable at b as the curve is discontinuous at this point.

c

y

x

The curve is not differentiable at point c since it is not smooth at that point.

ch8.indd 450 7/10/09 12:00:45 PM


EXAMPLES

1. Find all points where the function below is not differentiable.

C

B

A

y

x

Solution

The function is not differentiable at points A and B since there are sharp corners and the curve is not smooth at these points.

It is not differentiable at point C since the function is discontinuous at this point.

2. Is the function ( )f xx x

x x1

3 2 1for

for 2

1$

=-

) differentiable at all points?

Solution

The functions ( ) ( ) 3 2f x x f x xand2= = - are both differentiable at all points. However, we need to look at where one fi nishes and the other starts, at f (1).

f

f

( )

( )

f x x

f x x

1 11

3 2

1 3 1 21

For

For

2

2

=

=

=

= -

= -

=

]

] ]

g

g g

This means that both pieces of this function join up (the function is continuous). However, to be differentiable, the curve must be smooth at this point.

CONTINUED

ch8.indd 451 6/29/09 12:53:43 PM


8.2 Exercises

For each function, state whether it has any points at which it is not differentiable.

Sketching this function shows that it is not smooth (it has a sharp corner) so it is not differentiable at 1x = .

1

1

-2

y = x2

y = 3x - 2

y

x

1. y

x

2.

x1

y

x

3.

x1

y

x

4. y

x

ch8.indd 452 6/29/09 12:53:55 PM


5.

x1 x2

y

x

6. ( ) 4f x x=

7. 3

1yx

= -+

8. ( )f xx xx x

21 2

if if

3 2#

=+

)

9. ( )f x

x x

x

x x

2 3

3 2 3

1 2

for

for

for

2

2

1

# #= -

- -

Z

[

\

]]

]

10.

-4

-4

-5

-3

-3

-2

-2

-1-1

2

1

3

4

5

1 2 3 4x

y

11. tany x= for x0 360c c# #

12. ( )f x xx

=

13. ( ) cosf 3 2i i= -

14. ( ) sing 2z z=

15. 93y

xx

2=

-

-

Limits

To differentiate from fi rst principles, we need to look more closely at the concept of a limit.

A limit is used when we want to move as close as we can to something. Often this is to fi nd out where a function is near a gap or discontinuous point. You saw this in Chapter 5 when looking at discontinuous graphs. In this topic, it is used when we want to move from a gradient of a line between two points to a gradient of a tangent.

EXAMPLES

1. Find .limx

x x2

2x 2

2

-

- -"

Solution

( )

( ) ( )

( 1)

lim lim

limx

x xx

x x

x2

22

1 2

2 1

3

x x

x

2

2

2

2

-

- -=

-

+ -

= +

= +

=

" "

"

You did this in Chapter 5.

CONTINUED

ch8.indd 453 7/10/09 12:01:00 PM


2. Find an expression in terms of x for 2 3limh

xh h h0

2

h

- -"

.

Solution

( )

( )

lim lim

limh

xh h hh

h x h

x h

x

2 3 2 3

2 3

2 3

h h

h

0

2

0

0

- -=

- -

= - -

= -

" "

"

3. Find an expression in terms of x for 3 5limx

x x x x0

2 2

x dd d d+ -

"d .

Solution

( )

( )

lim lim

limx

x x x xx

x x x

x x

x

3 5 3 5

3 5

3 5

x x

x

0

2 2

0

2

0

2

2

dd d d

d

d d

d

+ -=

+ -

= + -

= -

" "

"

d d

d

1. Evaluate

(a) 3lim xx x

0

2

x

+"

(b) 5 2 7lim xx x x

0

3 2

x

- -"

(c) 33lim

xx x

3

2

x -

-"

(d) 416lim

tt

4

2

t -

-"

(e) 11

limgg

1

2

g -

-

"

(f) 2

2limx

x x2

2

x +

+ -"-

(g) 2limh

h h0

5

h

+"

(h) 3

7 12limx

x x3

2

x -

- +"

(i) 525lim

nn

5

2

n -

-"

(j) 1

4 3limx

x x1 2

2

x -

+ +"-

2. Find as an expression in terms of x

(a) 2 4limh

x h xh h0

2

h

- -"

(b) 2limh

x h xh h0

3

h

+ -"

(c) 3 7 4limh

x h xh h h0

2 2 2

h

- + -"

(d) 4 4limh

x h x h xh0

4 2 2

h

- -"

(e) 3 4 3limh

x h xh xh h0

2 2 2

h

+ - +"

(f) 2 5 6limh

x h xh h0

2 2

h

+ +"

(g) 2limx

x x x x0

2 2

x dd d-

"d

(h) 4 2limx

x x x0

2 2

x dd d-

"d

(i) 3limx

x x x x x0

3 2

x dd d d+ -

"d

(j) 2 9limx

x x x x x0

2

x dd d d- +

"d

8.3 Exercises

ch8.indd 454 6/29/09 12:54:17 PM


Differentiation as a limit

The formula m x xy y

2 1

2 1=

-

- is used to fi nd the gradient of a straight line when we

know two points on the line. However, when the line is a tangent to a curve, we only know one point on the line—the point of contact with the curve.

To differentiate from fi rst principles , we fi rst use the point of contact and another point close to it on the curve (this line is called a secant ) and then we move the second point closer and closer to the point of contact until they overlap and the line is at single point (the tangent ). To do this, we use a limit.

If you look at a close up of a graph, you can get some idea of this concept. When the curve is magnifi ed, two points appear to be joined by a straight line. We say the curve is locally straight .

Investigation

Use a graphics calculator or a computer program to sketch a curve and then zoom in on a section of the curve to see that it is locally straight.

For example, here is a parabola.

-20

-10

2 20

10

2

y

xf1(x) = x2

Notice how it looks straight when we zoom in on a point on the parabola?

f 1(x) = x22.99

7.99 y

x

Use technology to sketch other curves and zoom in to show that they are locally straight.

ch8.indd 455 6/29/09 12:54:27 PM


Before using limits to fi nd different formulae for differentiating from fi rst principles, here are some examples of how we can calculate an approximate value for the gradient of the tangent to a curve. By taking two points close together, as in the example below, we fi nd the gradient of the secant and then estimate the gradient of the tangent.

(3, f (3))

(3.01, f (3.01))

y

x

EXAMPLES

1. For the function f x x3=] g , fi nd the gradient of the secant PQ where P is the point on the function where 2x = and Q is another point on the curve close to P . Choose different values for Q and use these results to estimate the gradient of the curve at P .

(2, f(2))

(2.1, f(2.1))y

Q

P

x

ch8.indd 456 7/25/09 5:34:01 PM


Solution

2, (2)P f= ^ h Take different values of x for point Q , for example 2.1x = Using different values of x for point Q gives the results in the table.

Point Q Gradient of secant PQ

. , .f2 1 2 1]_ gi.

( . ) ( )

..

.

mf f

2 1 22 1 2

2 1 22 1 2

12 61

3 3

=-

-

=-

-

=

. , .f2 01 2 01]_ gi.

( . ) ( )

..

.

mf f

2 01 22 01 2

2 01 22 01 2

12 0601

3 3

=-

-

=-

-

=

. , .f2 001 2 001]_ gi.

( . ) ( )

..

.

mf f

2 001 22 001 2

2 001 22 001 2

12 006001

3 3

=-

-

=-

-

=

. , .f1 9 1 9]_ gi.

( . ) ( )

..

.

mf f

1 9 21 9 2

1 9 21 9 2

11 41

3 3

=-

-

=-

-

=

. , .f1 99 1 99]_ gi.

( . ) ( )

..

.

mf f

1 99 21 99 2

1 99 21 99 2

11 9401

3 3

=-

-

=-

-

=

. , .f1 999 1 999]_ gi.

( . ) ( )

..

.

mf f

1 999 21 999 2

1 999 21 999 2

11 994001

3 3

=-

-

=-

-

=

From these results, a good estimate for the gradient at P is 12. We can say that as x approaches 2, the gradient approaches 12.

We can write 2

( ) (2)12lim

x

f x f2x -

-=

" .

Use y

mx x

y

2 1

12

-

-= to fi nd

the gradient of the secant.

CONTINUED

ch8.indd 457 6/29/09 12:54:47 PM


2. For the curve y x2= , fi nd the gradient of the secant AB where A is the point on the curve where 5x = and point B is close to A . Find an estimate of the gradient of the curve at A by using three different values for B .

Solution

5, (5)A f= ^ h Take three different values of x for point B , for example . , .x x4 9 5 1= = and 5.01x = .

(a) . , ( . )

.( . ) ( )

..

.

B f

m x xy y

f f

4 9 4 9

4 9 54 9 5

4 9 54 9 5

9 9

2 1

2 1

2 2

=

=-

-

=-

-

=-

-

=

^ h

(b) . , ( . )

.( . ) ( )

..

.

B f

m x xy y

f f

5 1 5 1

5 1 55 1 5

5 1 55 1 5

10 1

2 1

2 1

2 2

=

=-

-

=-

-

=-

-

=

^ h

(c) . , ( . )

.( . ) ( )

..

.

B f

m x xy y

f f

5 01 5 01

5 01 55 01 5

5 01 55 01 5

10 01

2 1

2 1

2 2

=

=-

-

=-

-

=-

-

=

^ h

From these results, a good estimate for the gradient at A is 10. We can say that as x approaches 5, the gradient approaches 10.

We can write 5

( ) (5)10lim

xf x f

5x -

-=

" .

We can fi nd a general formula for differentiating from fi rst principles by using c rather than any particular number. We use general points , ( )P c f c^ h and , ( )Q x f x^ h where x is close to c .

The gradient of the secant PQ is given by

( ) ( )

m x xy y

x cf x f c

2 1

2 1=

-

-

=-

-

ch8.indd 458 6/29/09 12:54:56 PM


The gradient of the tangent at P is found when x approaches c . We call this ( )cfl .

( )( ) ( )

limf c x cf x f c

x c=

-

-

"l

There are other versions of this formula. We can call the points , ( )P x f x^ h and , ( )Q x h f x h+ +^ h where h is small.

(x + h, f(x + h))

(x , f(x))P

Q

y

x

Secant PQ has gradient

( ) ( )

( ) ( )

m x xy y

x h x

f x h f x

h

f x h f x

2 1

2 1=

-

-

=+ -

+ -

=+ -

To fi nd the gradient of the secant, we make h smaller as shown, so that Q becomes closer and closer to P .

P

Q

Q

Q

Q

y

x

(x + h, f(x + h))

(x, f (x))

Search the Internet using keywords ‘differentiation from

fi rst principles’, gradient of secant’ and ‘tangent’ to fi nd mathematical websites that

show this working.

ch8.indd 459 6/29/09 12:55:04 PM


As h approaches 0, the gradient of the tangent becomes ( ) ( )

limh

f x h f x0h

+ -

" .

We call this ( )xfl .

( )( ) ( )

limxh

f x h f xh 0

=+ -

"fl

If we use ,P x y^ h and ,Q x x y yd d+ +^ h close to P where x yandd d are small:

Gradient of secant PQ

m x xy

x x x

y y y

x

y

y

2 1

2 1

d

d

d

d

=-

-

=+ -

+ -

=

As xd approaches 0, the gradient of the tangent becomes limx

yx 0 d

d

"d . We

call this dx

dy .

The symbol d is a Greek letter called delta.

limdx

dy

x

yx 0d

d=

"d

All of these different notations stand for the derivative, or the gradient of the tangent:

, ( ), ( ) , ( ),dx

dy

dxd y

dxd f x f x yl l^ h

These occur because Newton, Leibniz and other mathematicians over the years have used different notation.

Investigation

Leibniz used dx

dy where d stood for ‘difference’. Can you see why he would

have used this?

Use the Internet to explore the different notations used in calculus and where they came from.

ch8.indd 460 6/29/09 12:55:16 PM


The three formulae for differentiating from fi rst principles all work in a similar way.

EXAMPLE

Differentiate from fi rst principles to fi nd the gradient of the tangent to the curve 3y x2= + at the point where 1.x =

Solution

Method 1:

( )( ) ( )

( )( ) ( )

( )( ) ( )

( )

( ) ( )

( )

lim

lim

lim

lim

lim

lim

lim

f c x cf x f c

f x x

f

f c x cf x f c

fx

f x f

xx

xx

xx x

x

3

1 1 34

11

1

13 4

11

11 1

1

1 1

2

x c

x c

x

x

x

x

x

2

2

1

1

2

1

2

1

1

=-

-

= +

= +

=

=-

-

=-

-

=-

+ -

=-

-

=-

+ -

= +

= +

=

"

"

"

"

"

"

"

l

l

l

]]gg

Method 2:

( )( ) ( )

limf xh

f x h f x

f x x

f

f x h x hx

f h h

h h

h h

3

1 1 34

31

1 1 3

1 2 3

2 4

When

h 0

2

2

2

2

2

2

=+ -

= +

= +

=

+ = + +

=

+ = + +

= + + +

= + +

"l

]]

] ]

] ]

gg

g g

g g

Remember that 3y x2= - is the same as ( ) .f x x 32= -

CONTINUED

ch8.indd 461 6/29/09 12:55:26 PM


( )( ) ( )

( )( ) ( )

( )

( )

( )

lim

lim

lim

lim

lim

lim

f xh

f x h f x

fh

f h f

hh h

hh h

hh h

h

11 1

2 4 4

2

2

2

2 0

2

h

h

h

h

h

h

0

0

0

2

0

2

0

0

=+ -

=+ -

=+ + -

=+

=+

= +

= +

=

"

"

"

"

"

"

l

l

Method 3:

lim

dx

dy

x

y

y x 3

x 0

2

d

d=

= +

"d

1 3

x

y

1

4

When2

=

= +

=

So point ,1 4^ h lies on the curve. Substitute point ( , )x y1 4d d+ + :

( )

( )

( )

lim

lim

y x

x x

x x

y x x

x

y

xx x

x

x x

x

dx

dy

x

y

x

4 1 3

1 2 3

2 4

2

2

2

2

2

2 0

2

x

x

2

2

2

2

2

0

0

d d

d d

d d

d d d

d

d

dd d

d

d d

d

d

d

d

+ = + +

= + + +

= + +

= +

=+

=+

= +

=

= +

= +

=

"

"

d

d

We can also use these formulae to fi nd the derivative function generally.

ch8.indd 462 7/25/09 4:23:38 AM


EXAMPLE

Differentiate f x 2x x2 7 3= + -] g from fi rst principles.

Solution

f x x x

f x h x h x h

x xh h x hx xh h x h

2 7 3

2 7 3

2 2 7 7 32 4 2 7 7 3

2

2

2 2

2 2

= + -

+ = + + + -

= + + + + -

= + + + + -

]] ] ]

^

gg g g

h

f x h f x x xh h x h x xx xh h x h x x

xh h h

2 4 2 7 7 3 2 7 32 4 2 7 7 3 2 7 3

4 2 7

2 2 2

2 2 2

2

+ - = + + + + - - + -

= + + + + - - - +

= + +

] ] ^ ^g g h h

( )( ) ( )

( )

( )

lim

lim

lim

lim

f xh

f x h f x

hxh h h

hh x h

x h

x

x

4 2 7

4 2 7

4 2 7

4 0 7

4 7

h

h

h

h

0

0

2

0

0

=+ -

=+ +

=+ +

= + +

= + +

= +

"

"

"

"

l

Try this example using the other two formulae.

1. (a) Find the gradient of the secant between the point ,1 2^ h and the point where 1.01x = , on the curve 1.y x4= +

Find the gradient of the (b) secant between ,1 2^ h and the point where 0.999x = on the curve.

Use these results to fi nd the (c) gradient of the tangent to the curve 1y x4= + at the point ,1 2^ h .

2. A function f x x x3= +] g has a tangent at the point ,2 10^ h .

Find the value of (a) 2

( ) (2)x

f x f

-

-

when 2.1x = .

Find the value of (b) ( ) ( )

x

f x f

22

-

-

when 2.01x = .

Evaluate (c) 2

( ) (2)x

f x f

-

- when

.x 1 99= . Hence fi nd the gradient of the (d)

tangent at the point ,2 10^ h . 3. For the function ,f x x 42= -] g

fi nd the derivative at point P where 3x = by selecting points near P and fi nding the gradient of the secant.

4. If ( )f x x2= , fi nd (a) ( )f x h+

show that (b) ( ) ( )f x h f x+ - xh h2 2= +

8.4 Exercises

ch8.indd 463 6/29/09 12:55:47 PM


show that (c)

( ) ( )

h

f x h f x+ - x h2= +

show that (d) ( )x x2=fl .

5. A function is given by ( ) 2 7 3f x x x2= - + .

Show that (a) ( )f x h+ =

2 4 2 7 7 3x xh h x h2 2+ + - - + . (b) Show that

( ) ( ) 4 2 7f x h f x xh h h2+ - = + - . Show that (c)

( ) ( )

4 2 7h

f x h f xx h

+ -= + - .

Find (d) ( )xfl .

6. A function is given by ( ) 5f x x x2= + + .

Find (a) f 2] g . Find (b) f h2 +] g . Find (c) f h f2 2+ -] ]g g . Show that (d)

( ) ( )

5h

f h fh

2 2+ -= + .

Find (e) (2)fl .

7. Given the curve ( ) 4 3f x x3= - fi nd (a) f 1-] g fi nd (b) f h1 1- + -f-] ]g g fi nd the gradient of the (c)

tangent to the curve at the point where x 1= - .

8. For the parabola 1y x2= - fi nd (a) f 3] g fi nd (b) f h f3 3+ -] ]g g fi nd (c) (3)fl .

9. For the function ( )f x x x4 3 5 2= - -

fi nd (a) ( )f 1l similarly, fi nd the gradient (b)

of the tangent at the point , 2 10- -^ h .

10. For the parabola y x x22= + show that (a)

2 2y x x x x2d d d d= + +

by substituting the point ,x x y yd d+ +^ h

show that (b) 2 2x

yx x

d

dd= + +

fi nd (c) dx

dy .

11. Differentiate from fi rst principles to fi nd the gradient of the tangent to the curve

(a) f x x2=] g at the point where 1x =

(b) y x x2= + at the point ,2 6^ h (c) f x x2 52= -] g at the point

where x 3= - (d) 3 3 1y x x2= + + at the point

where 2x = (e) f x x x7 42= - -] g at the

point , 1 6-^ h . 12. Find the derivative function for

each curve by differentiating from fi rst principles

(a) f x x2=] g (b) y x x52= + (c) f x x x4 4 32= - -] g (d) y x x5 12= - - (e) y x3= (f) f x x x2 53= +] g (g) 2 3 1y x x x3 2= - + - (h) ( )f x x2 3= - .

13. The curve y x= has a tangent drawn at the point ,4 2^ h .

Evaluate (a) 4

( ) (4)x

f x f-

- when

.x 3 9= . Evaluate (b)

4( ) (4)

xf x f

-

- when

.x 3 999= . Evaluate (c)

4( ) (4)

xf x f

-

- when

4.01x = .

14. For the function ( )f x x 1= - ,

evaluate (a) 5

( ) (5)x

f x f-

- when

.x 4 99= .

Remember that

xx

11=

-

ch8.indd 464 6/29/09 12:55:58 PM


evaluate (b) ( ) ( )x

f x f5

5-

- when

5.01x = . Use these results to fi nd the (c)

derivative of the function at the point where x 5= .

15. Find the gradient of the tangent

to the curve 4yx2

= at point

,P 2 1^ h by fi nding the gradient of the secant between P and a point close to P .

Short Methods of Differentiation

The basic rule

Remember that the gradient of a straight line y mx b= + is m . The tangent to the line is the line itself, so the gradient of the tangent is m everywhere along the line.

y

x

y = mx + b

So if ,y mxdx

dym= =

For a horizontal line in the form y k= , the gradient is zero.

y

x

y = k

So if ,y kdx

dy0= =

dxd kx k=] g

dxd k 0=] g

ch8.indd 465 7/25/09 4:23:39 AM


Proof

Investigation

Differentiate from fi rst principles:

y x

y x

y x

2

3

4

=

=

=

Can you fi nd a pattern? Could you predict what the result would be for x n ?

Alternatively, you could fi nd an approximation to the derivative of a

function at any point by drawing the graph of .

( . ) ( )y

f x f x

0 010 01

=+ -

.

Use a graphics calculator or graphing computer software to sketch the derivative for these functions and fi nd the equation of the derivative.

Mathematicians working with differentiation from fi rst principles discovered this pattern that enabled them to shorten differentiation considerably!

For example: When ,y x y x22= =l When ,y x y x33 2= =l When ,y x y x44 3= =l

dxd x nxn n 1= -^ h

You do not need to know this proof.

( )

( ) ( )

( ) ( ) ( )

( ) [( ) ( ) ( ) ( )

. . . ( ) ]

[( ) ( ) ( ) ( )

. . . ( ) ]

f x x

f x h x h

f x h f x x h x

x h x x h x h x x h x x h x

x h x x

h x h x h x x h x x h x

x h x x

n

n

n n

n n n n

n n

n n n n

n n

1 2 3 2 4 3

2 1

1 2 3 2 4 3

2 1

=

+ = +

+ - = + -

= + - + + + + + + +

+ + + +

= + + + + + + +

+ + + +

- - - -

- -

- - - -

- -

^ h

( )( ) ( )

[( ) ( ) ( ) ( ) . . . ( ) ]

[( ) ( ) ( ) ( ) . . . ( ) ]

( ) ( ) ( ) ( ) . . . ( )

lim

lim

lim

f xh

f x h f x

hh x h x h x x h x x h x x h x x

x h x h x x h x x h x x h x x

x x x x x x x x x x

nx

h

h

n n n n n n

h

n n n n n n

n n n n n n

n

0

0

1 2 3 2 4 3 2 1

0

1 2 3 2 4 3 2 1

1 2 3 2 4 3 2 1

1

=+ -

=+ + + + + + + + + + +

= + + + + + + + + + + +

= + + + + + +

=

"

"

"

- - - - - -

- - - - - -

- - - - - -

-

l

ch8.indd 466 6/29/09 12:56:18 PM



EXAMPLE

Differentiate ( )f x x7= .

Solution

( ) 7f x x6=l

There are some more rules that give us short ways to differentiate functions. The fi rst one says that if there is a constant in front of the x (we call this a

coeffi cient), then it is just multiplied with the derivative.

dxd kx knxn n 1= -^ h

( ) ( )dxd kf x kf x= l^ h

A more general way of writing this rule is:

Proof

( )( ) ( )

[ ( ) ( )]

( ) ( )

( )

lim

lim

lim

dxd kf x

h

kf x h kf x

h

k f x h f x

kh

f x h f x

kf x

h

h

h

0

0

0

=+ -

=+ -

=+ -

=

"

"

"

l

^ h

EXAMPLE

Find the derivative of 3 x 8 .

Solution

3y x

dx

dyx

x

3 8

24

If 8

7

7

#

=

=

=

ch8.indd 467 6/29/09 12:56:28 PM


Many functions use a combination of these rules.

Also, if there are several terms in an expression, we differentiate each one separately. We can write this as a rule:

( ) ( ) ( ) ( )dxd f x g x f x g x+ = + ’l^ h

Proof

( ) ( )[ ( ) ( )] [ ( ) ( )]

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

lim

lim

lim

lim

lim lim

dxd f x g x

h

f x h g x h f x g x

h

f x h g x h f x g x

h

f x h f x g x h g x

h

f x h f x

h

g x h g x

h

f x h f x

h

g x h g x

f x g x

h

h

h

h

h h

0

0

0

0

0 0

+ =+ + + - +

=+ + + - -

=+ - + + -

=+ -

++ -

=+ -

++ -

= +

"

"

"

"

" "

l l

^ h

= G


EXAMPLE

Differentiate x x3 4+ .

Solution

( )dxd x x x x3 43 4 2 3+ = +

EXAMPLES

Differentiate 1. 7 x

Solution

dxd x7 7=] g

CONTINUED

ch8.indd 468 6/29/09 12:56:37 PM


2. ( ) 5x x xf 4 3= - +

Solution

( )f x x x

x x

4 3 0

4 3

3 2

3 2

= - +

= -

l

3. y x4 7=

Solution

dx

dyx

x

4 7

28

6

6

#=

=

4. If ( )f x x x x2 7 5 45 3= - + - , evaluate ( )f 1-l

Solution

( )

( ) ( ) ( )

f x x x

f

10 21 5

1 10 1 21 1 56

4 2

4 2

= - +

- = - - - +

= -

l

l

5. Differentiate 2

3 5x

x x2 +

Solution

Divide by 2 x before differentiating.

xx x

xx

xx

x

dx

dy

23 5

23

25

23

25

23

121

2 2+= +

= +

=

=

6. Differentiate S r rh2 22r r= + with respect to r .

Solution

We are differentiating with respect to r , so r is the variable and r and h are constants.

( )

drdS r h

r h

2 2 2

4 2

r r

r r

= +

= +

ch8.indd 469 6/29/09 12:56:46 PM


1. Differentiate (a) 2x + (b) 5 9x - (c) 3 4x x2 + + (d) 5 8x x2 - - (e) 2 7 3x x x3 2+ - - (f) 2 7 7 1x x x3 2- + - (g) 3 2 5x x x4 2- + (h) 5 2x x x6 5 4- - (i) 2 4 2 4x x x x5 3 2- + - + (j) 4 7x x10 9-

2. Find the derivative of (a) 2 1x x +] g (b) 2 3x 2-] g (c) 4 4x x+ -] ]g g (d) 2 3x2 2

-^ h (e) 2 5 1x x x2+ - +] ^g h

3. Differentiate

(a) x x6

2

-

(b) 2 3

4x x4 3

- +

(c) ( )x x31 36 2 -

(d) xx x2 53 +

(e) 4

2x

x x2 +

(f) 3

2 3 6 2x

x x x x2

5 4 3 2- + -

4. Find ( )f xl when ( )f x x8 2= - x7 4+ .

5. If y x x2 54 3= - + , fi nd dx

dy when

.x 2= -

6. Find dx

dy if

y x x x6 5 710 8 5= - + - .x3 8+

7. If s t t5 202= - , fi nd dtds .

8. Find ( )g xl given ( )g x x5 4= - .

9. Find dtdv when v t15 92= - .

10. If h t t40 2 2= - , fi nd dtdh .

11. Given 34V r3r= , fi nd

drdV .

12. If ( )f x x x2 3 43= - + , evaluate (1)fl .

13. Given ( ) 5f x x x2= - + , evaluate (a) ( )f 3l (b) ( )f 2-l (c) x when ( )f x 7=l

14. If y x 73= - , evaluate

(a) dx

dy when x 2=

(b) x when dx

dy12=

15. Evaluate ( )g 2l when

( )g t t t t3 4 2 13 2= - - + .

8.5 Exercises

Expand brackets before differentiating.

Simplify by dividing before differentiating.

ch8.indd 470 6/29/09 12:56:55 PM


DID YOU KNOW?

• The word tangent comes from the Latin ‘tangens’, meaning ‘touching’. A tangent to a circle intersects it only once.

• However, a tangent to a curve could intersect the curve more than once.

• A line may only intersect a curve once but not be a tangent.

• So a tangent to a curve is best described as the limiting position of the secant PQ as Q approaches P .

This line is a tangent to the curve at point P.

Remember from earlier in the chapter that the derivative is the gradient of the tangent to a curve.

dx

dy is the gradient of the tangent to a curve

Tangents and Normals

ch8.indd 471 7/25/09 4:23:40 AM


u EXAMPLES

1. Find the gradient of the tangent to the parabola y x 12= + at the point ,1 2^ h . Solution

, ( )

dx

dyx

x

dx

dy

2 0

2

1 2 2 1

2

At

= +

=

=

=

^ h

So the gradient of the tangent at ,1 2^ h is 2.

2. Find values of x for which the gradient of the tangent to the curve 2 6 1y x x3 2= - + is equal to 18.

Solution

dx

dyx x6 122= -

dx

dy is the gradient of the tangent, so substitute

dx

dy18= .

x 1= -

,

,

x x

x x

x xx x

x x

x

18 6 12

0 6 12 18

2 33 1

3 0 1 0

3

2

2

2

`

= -

= - -

= - -

= - +

- = + =

=

] ]g g

3. Find the equation of the tangent to the curve y x x x3 7 24 3= - + - at the point ,2 4^ h . Solution

,

dx

dyx x

dx

dy

4 9 7

2 4 4 2 9 2 7

3

At

3 2

3 2

= - +

= - +

=

^ ] ]h g g

So the gradient of the tangent at ,2 4^ h is 3. Equation of the tangent:

y y m x x

y x4 3 21 1- = -

- = -

_]

ig

ch8.indd 472 6/29/09 12:57:14 PM


x

y x

x y

3 6

3 2

0 3 2or

= -

= -

= - -

The normal is a straight line perpendicular to the tangent at the same point of contact with the curve.

Normal

Tangent

y

x

If lines with gradients m 1 and m 2 are perpendicular, then m m 11 2 = - You used this rule in the

previous chapter.

EXAMPLES

1. Find the gradient of the normal to the curve y x x2 3 52= - + at the point where x 4= .

Solution

dx

dy is the gradient of the tangent .

13=

x4 3= -dx

dy

x

dx

dy

m

4

4 4 3

13

When

So 1

#

=

= -

=

The normal is perpendicular to the tangent . m m 1So 1 2 = -

CONTINUED

ch8.indd 473 6/29/09 12:57:23 PM


m

m

13 1

131

2

2

= -

= -

So the gradient of the normal is 131

- .

2. Find the equation of the normal to the curve y x x x3 2 13 2= + - - at the point , .1 3-^ h

Solution

dx

dy is the gradient of the tangent .

dx

dyx x

x

dx

dy

m

3 6 2

1

3 1 6 1 2

5

5

When

So

2

2

1

= + -

= -

= - + - -

= -

= -

] ]g g

The normal is perpendicular to the tangent .

m m

m

m

1

5 1

51

So 1 2

2

2

= -

- = -

=

So the gradient of the normal is 51 .

Equation of the normal:

y y m x x

y x

y x

x y

351 1

5 15 1

0 5 16

1 1- = -

- = - -

- = +

= - +

_]]igg

1. Find the gradient of the tangent to the curve

(a) y x x33= - at the point where 5x =

(b) f x 2x x 4= + -] g at the point ,7 38-^ h

(c) f x 3x x5 4 1= - -] g at the point where x 1= -

(d) 5 2 3y x x2= + + at the point ,2 19-^ h

(e) y x2 9= at the point where 1x =

(f) f x 3x 7= -] g at the point where 3x =

(g) v t t2 3 52= + - at the point where t 2=

(h) 3 2 8 4Q r r r3 2= - + - at the point where 4r =

(i) 4h t t4= - where t 0= (j) f t t t t3 8 55 3= - +] g at the

point where 2t = .

8.6 Exercise s

ch8.indd 474 6/29/09 12:57:33 PM


2. Find the gradient of the normal to the curve

(a) f x x x2 2 13= + -] g at the point where x 2= -

(b) 3 5 2y x x2= + - at the point ,5 48-^ h

(c) f x x x2 72= - -] g at the point where 9x = -

(d) 3 2y x x x3 2= + + - at the point ,4 62- -^ h

(e) f x x10=] g at the point where 1x = -

(f) 7 5y x x2= + - at the point ,7 5- -^ h

(g) 2 3 1A x x x3 2= + - + at the point where 3x =

(h) f a a a3 2 62= - -] g at the point where 3a = -

(i) 4 9V h h3= - + at the point ,2 9^ h

(j) 2 5 3g x x x x4 2= - + -] g at the point where x 1= - .

3. Find the gradient of the (i) tangent and (ii) normal to the curve

(a) 1y x2= + at the point ,3 10^ h (b) f x x5 2= -] g at the point

where x 4= - (c) 2 7 4y x x5 2= - + at the point

where x 1= - (d) 3 2 8p x x x x6 4= - - +] g

where 1x = (e) f x x x4 2= - -] g at the point

,6 26-^ h . 4. Find the equation of the tangent

to the curve (a) 5 1y x x4= - + at the

point ,2 7^ h (b) ( ) 5 3 2 6f x x x x3 2= - - + at

the point ,1 6^ h (c) 2 8y x x2= + - at the

point ,3 5- -^ h (d) 3 1y x3= + at the point

where 2x = (e) 4 7 2v t t4 3= - - at the point

where 2t =

5. Find the equation of the normal to the curve

(a) f x x x3 53= - +] g at the point ,3 23^ h

(b) 4 5y x x2= - - at the point ,2 7-^ h

(c) f x x x7 2 2= -] g at the point where 6x =

(d) 7 3 2y x x2= - - at the point ,3 70-^ h

(e) 2 4 1y x x x4 3= - + + at the point where 1x = .

6. Find the equation of the (i) tangent and (ii) normal to the curve

(a) f x x x4 82= - +] g at the point ,1 11^ h

(b) 2 5y x x x3 2= + - at the point ,3 6-^ h

(c) 5F x x x5 3= -] g at the point where 1x =

(d) 8 7y x x2= - + at the point ,3 8-^ h

(e) 2 4 1y x x x4 3= - + + at the point where 1x = .

7. For the curve 27 5,y x x3= - -

fi nd values of x for which 0dx

dy= .

8. Find the coordinates of the point at which the curve 1y x3= + has a tangent with a gradient of 3.

9. A function ( ) 4 12f x x x2= + - has a tangent with a gradient of 6- at point P on the curve. Find the coordinates of the point P .

10. The tangent at point P on the curve 4 1y x2= + is parallel to the x -axis. Find the coordinates of P .

11. Find the coordinates of point Q where the tangent to the curve 5 3y x x2= - is parallel to the line 7 3 0x y- + = .

ch8.indd 475 6/29/09 12:57:43 PM


12. Find the coordinates of point S where the tangent to the curve 4 1y x x2= + - is perpendicular to the line 4 2 7 0x y+ + = .

13. The curve 3 4y x2= - has a gradient of 6 at point A .

Find the coordinates of (a) A . Find the equation of the (b)

tangent to the curve at A .

14. A function 3 2 5h t t2= - + has a tangent at the point where t 2= . Find the equation of the tangent.

15. A function f x x x2 8 32= - +] g has a tangent parallel to the line 4 2 1 0x y- + = at point P . Find the equation of the tangent at P .

Further Differentiation and Indices

The basic rule for differentiating x n works for any rational number n .

Investigation

(a) Show that 1. 1 1( )x h x x x h

h+

- =+

- .

Hence differentiate (b) 1y x= from fi rst principles .

Differentiate (c) y x 1= - using a short method. Do you get the same answer as 1(b)? (a) Show that 2. ( )( )x h x x h x h+ - + + = .

Hence differentiate (b) y x= from fi rst principles .

Differentiate (c) 2y x=1

and show that this gives the same answer as 2(b).

EXAMPLES

1. Differentiate 7 x3 .

Solution

3

x x

dx

dyx

x

x

7 7

731

37

37 1

3 3

31

3

2

$

#

=

=

=

=

-

-

1

1

2

We sometimes need to change a function into index form before differentiating.

ch8.indd 476 6/29/09 12:57:52 PM


x

x

37 1

3

7

23

23

#=

=

2. Find the equation of the tangent to the curve 4yx2

= at the point where 2.x =

Solution

yxx

dx

dyx

x

4

4

8

8

2

2

3

3

=

=

= -

= -

-

-

2xWhen =

y24

1

2=

=

Gradient of the tangent at 2,1 :^ h dx

dy

28

1

3= -

= -

Equation of the tangent:

y y m x x

y xx

y x

x y

1 1 22

3

3 0or

1 1- = -

- = - -

= - +

= - +

+ - =

_]

ig

8.7 Exercises

1. Differentiate

(a) x 3- (b) x1.4 (c) 6x0.2

(d) 2x1

(e) 22 3x x 1- -1

(f) 3x 31

(g) 8x 4

3

(h) 22x--

1

2. Find the derivative function, writing the answer without negative or fractional indices .

(a) 1x

(b) 5 x (c) x6

(d) 2x5

(e) 5x3

-

(f) 1x

ch8.indd 477 6/29/09 12:58:01 PM


(g) x21

6

(h) x x

(i) x3

2

(j) x x41 3

2 4+

3. Find the gradient of the tangent to the curve y x3= at the point where 27.x =

4. If 12xt

= , fi nd dtdx when .t 2=

5. A function is given by ( )f x x4= . Evaluate .( )f 16l

6. Find the gradient of the tangent

to the curve 23yx2

= at the point

1, 121c m .

7. Find dx

dy if y x x

2= +^ h .

8. A function ( )2

f xx

= has a

tangent at , .4 1^ h Find the gradient of the tangent.

9. Find the equation of the tangent

to the curve 1yx3

= at the point

2,81c m .

10. Find the equation of the tangent to ( ) 6f x x= at the point where 9.x =

11. (a) Differentiate xx

.

H(b) ence fi nd the gradient of the

tangent to the curve y xx

= at

the point where 4.x =


to the curve 4y x= at the point

8,21c m .

13. If the gradient of the tangent to

y x= is 61 at point A , fi nd the

coordinates of A .

14. The function ( ) 3f x x= has

( )f x43

=l . Evaluate x .

15. The hyperbola 2y x= has two

tangents with gradient 252

- . Find

the coordinates of the points of contact of these tangents.

Note that 1

.x x2 2

1 16 6

#=

Use index laws to simplify fi rst.

Expand brackets fi rst.

Composite Function Rule

A composite function is a function composed of two or more other functions. For example, 3 4x2 5

-^ h is made up of a function u 5 where 3 4u x2= - . To differentiate a composite function, we need to use the result. .

This rule is also called the function of a function rule or chain rule.

dx

dy

du

dy

dxdu

#=

ch8.indd 478 7/10/09 12:01:22 PM


Proof

Let ,x yd d and ud be small changes in x , y and u where , , .x y u0 0 0" " "d d d

0, 0

lim lim lim

x

y

u

y

xu

x u

x

y

u

y

xu

Then

As

Sox u x0 0 0

" "

#

#

d

d

d

d

dd

d d

d

d

d

d

dd

=

=" " "d d d

Using the defi nition of the derivative from fi rst principles, this gives

dx

dy

du

dy

dxdu

#= .

You do not need to learn this proof.

EXAMPLES

Differentiate 1. (5 4)x 7+

Solution

Let 5 4u x= +

` 7

( )

dxdu

y u

du

dyu

dx

dy

du

dy

dxdu

u

x

5

7 5

35 5 4

Then

7

6

6

6

#

#

=

=

=

=

=

= +

2. (3 2 1)x x2 9+ -

Solution

( )

( ) ( )

u x x

dxdu x

y u

du

dyu

dx

dy

du

dy

dxdu

u x

x x x

3 2 1

6 2

9

9 6 2

9 6 2 3 2 1

Let

Then

2

9

8

8

2 8

#

= + -

= +

=

=

=

= +

= + + -

Can you see a quick way of doing this

question?

CONTINUED

ch8.indd 479 6/29/09 12:58:20 PM


3. 3 x-

Solution

2

2

2

2

2

( )

( 1)

(3 )

x xu x

dxdu

y u

du

dyu

dx

dy

du

dy

dxdu

u

x

x

3 33

1

21

21

21

2 31

Let

#

- = -

= -

= -

=

=

=

= -

= - -

= --

-

-

-

1

1

1

1

1

The derivative of a composite function is the product of two derivatives. One is the derivative of the function inside the brackets. The other is the derivative of the whole function.

[dxd n n 1-( )] ( ) [f x f x n= ( )]f xl

Proof

( )

( )

( )

( ) [ ( )]

u f x

dxdu f x

y u

du

dynu

dx

dy

du

dy

dxdu

nu f x

f x n f x

Let

Then

n

n

n

n

1

1

1

`

#

#

=

=

=

=

=

=

=

-

-

-

l

l

l


ch8.indd 480 6/29/09 1:01:44 PM


EXAMPLES

Differentiate 1. (8 1)x3 5-

Solution

( ) [ ( )]

( )

( )

dx

dyf x n f x

x x

x x

24 5 8 1

120 8 1

n 1

2 3 4

2 3 4

$

$

=

= -

= -

-l

2. (3 8)x 11+

Solution

.( ) [ ( )]

( )

( )

y f x n f x

x

x

3 11 3 8

33 3 8

n 1

10

10

#

=

= +

= +

-l l

3. (6 1)

1x 2+

Solution

( )( )

( ) [ ( )]

( )

( )

( )

xx

y f x n f x

x

x

x

6 11 6 1

6 2 6 1

12 6 1

6 112

n

22

1

3

3

3

$

#

+= +

=

= - +

= - +

= -+

-

-

-

-

l l

1. Differentiate (a) ( )x 3 4+ (b) ( )x2 1 3- (c) ( )x5 42 7- (d) ( )x8 3 6+ (e) ( )x1 5-

(f) 3(5 9)x 9+ (g) ( )x2 4 2- (h) ( )x x2 33 4+ (i) ( 5 1)x x2 8+ - (j) ( 2 3)x x6 2 6- +

(k) 2( )x3 1-1

8.8 Exercises

ch8.indd 481 8/1/09 10:51:00 PM


(l) (4 )x 2- -

(m) ( 9)x2 3- -

(n) 3( )x5 4+1

(o) 4( )x x x73 2- +3

(p) 3 4x +

(q) 5 2

1x -

(r) ( 1)

1x2 4+

(s) ( )x7 3 2-3

(t) 4

5x+

(u) 2 3 1

1x -

(v) 4(2 7)

3x 9+

(w) 3 31

x x x4 3- +

(x) ( )x4 1 43 +

(y) ( )x7

154 -

2. Find the gradient of the tangent to the curve 3 2y x 3= -] g at the point ., 1 1^ h

3. If ( ) 2( 3)f x x2 5= - , evaluate (2)fl .

4. The curve 3y x= - has a

tangent with gradient 21 at point

N . Find the coordinates of N .

5. For what values of x does the

function ( )4 1

1f xx

=-

have

( )f x494

= -l ?

6. Find the equation of the tangent to (2 1)y x 4= + at the point where 1.x = -

Product Rule

Differentiating the product of two functions y uv= gives the result

dx

dyudxdv v

dxdu

= +

Proof

y uv= Given that , y u vand d d d are small changes in y , u and v .

( ) ( )y y u u v v

uv u v v u u v

y u v v u u v y uv

x

yu

xv v

xu u

xv

since`

d d d

d d d d

d d d d d

d

d

dd

dd

ddd

+ = + +

= + + +

= + + =

= + +

^ h

ch8.indd 482 7/10/09 12:01:49 PM


0, 0

lim lim

lim lim lim

x u

x

yu

xv v

xu u

xv

uxv v

xu u

xv

dx

dyudxdv v

dxdu

As

x x

x x x

0 0

0 0 0

" "d d

d

d

dd

dd

ddd

dd

dd

ddd

= + +

= + +

= +

" "

" " "

d d

d d d

<< < <

FF F F

It is easier to remember this rule as y uv vu= +l l l . We can also write this the other way around which helps when learning the quotient rule in the next section.


If ,y uv y u v v u= = +l l l

EXAMPLES

Differentiate 1. 3 1 5x x+ -] ]g g Solution

You could expand the brackets and then differentiate:

x x x x x

x x

dx

dyx

3 1 5 3 15 5

3 14 5

6 14

2

2

+ - = - + -

= - -

= -

] ]g g

Using the product rule:

y uv u x v x

u v

3 1 5

3 1

where and= = + = -

= =l l

y u v v u

x xx x

x

3 5 1 3 13 15 3 1

6 14

= +

= - + +

= - + +

= -

l l l

] ]g g

2. 2 5 3x x5 3+] g Solution

.

y uv u x v x

u x v x

2 5 3

10 5 3 5 3

where and5 3

4 2

= = = +

= = +l l

]]gg

CONTINUED

ch8.indd 483 7/10/09 12:02:03 PM


.

y u v v u

x x x x

x x x x

x x x x

x x x

10 5 3 5 3 5 3 2

10 5 3 30 5 3

10 5 3 5 3 3

10 5 3 8 3

4 3 2 5

4 3 5 2

4 2

4 2

$

= +

= + + +

= + + +

= + + +

= + +

l l l

] ]] ]] ]] ]

g gg gg gg g6 @

3. (3 4) 5 2x x- -

Solution

2

2

2

( )

x x

y uv u x v x

u v x

5 2 5 2

3 4 5 2

3 221 5 2

Remember

where and

$

- = -

= = - = -

= = - --

1

1

1

l l

]]

gg

-

2

2

2

2

( )

( )

( )

( ) ( )

y u v v u

x x x

x x x

xx

x

xx

x

x

x x x

x

x x

xx x

xx

3 5 2 221 5 2 3 4

3 5 2 3 4 5 2

3 5 25 2

3 4

3 5 25 2

3 4

5 2

3 5 2 5 2 3 4

5 2

3 5 2 3 4

5 215 6 3 4

5 219 9

$

$

= +

= - +- - -

= - - - -

= - -

-

-

= - --

-

=-

- - - -

=-

- - -

=-

- - +

=-

-

-1 1

1

1

l l l

] ] ]]

g g gg

We can simplify this further by factorising.

1. Differentiate

(a) 2 3x x3 +] g (b) 3 2 2 1x x- +] ]g g (c) 3 5 7x x +] g (d) 4 3 1x x4 2 -^ h (e) 2 3x x x4 -^ h (f) 1x x2 3+] g

(g) 4 3 2x x 5-] g (h) x x3 44 3-] g (i) 1 2 5x x 4+ +] ]g g (j) 5 3 1x x x3 2 2 5

+ - +^ ^h h (k) 2x x-

(l) 2 15 3xx

-

+

8.9 Exercises

Change this into a product before differentiating.

ch8.indd 484 7/10/09 12:02:40 PM


2. Find the gradient of the tangent to the curve 2 3 2y x x 4= -] g at the point 1, 2^ h .

3. If ( ) (2 3)(3 1)f x x x 5= + - , evaluate ( )1fl .

4. Find the exact gradient of the tangent to the curve y x x2 5= + at the point where 1x = .

5. Find the gradient of the tangent where 3,t = given 2 5 1x t t 3= - +] ]g g .

6. Find the equation of the tangent to the curve 2 1y x x2 4= -] g at the point , .1 1^ h

7. Find the equation of the tangent to ( 1) ( 1)h t t2 7= + - at the point , .2 9^ h

8. Find exact values of x for which the gradient of the tangent to the curve 2 3y x x 2= +] g is 14.

9. Given ( ) (4 1)(3 2)f x x x 2= - + , fi nd the equation of the tangent at the point where x 1= - .

Quotient Rule

Differentiating the quotient of two functions y vu

= gives the result.

dx

dy

v

vdxdu u

dxdv

2=

-

Proof

y vu

=

Given that , y u vandd d d are small changes in y , u and v .

`

( )

( )

( )

( )

( )

( ) ( )

( )

( )

( ),

y yv vu u

yv vu u

vu y v

u

v v v

v u u

v v v

u v v

v v v

v u u u v v

v v vvu v u uv u v

v v vv u u v

x

y

v v v

vxu u

xv

x v0 0

since

As " "

ddd

ddd

d

d

d

d

d

d d

dd d

dd d

d

d

ddd

dd

d d

+ =+

+

=+

+- =

=+

+-

+

+

=+

+ - +

=+

+ - -

=+

-

=+

-

a k

ch8.indd 485 7/10/09 12:02:55 PM


( )lim lim

x

y

v v v

vxu u

xv

dx

dy

v

vdxdu u

dxdv

x x0 0

2

d

d

ddd

dd

=+

-

=

-

" "d d

R

T

SSSS

V

X

WWWW

It is easier to remember this rule as yv

u v v u2

=-

ll l

.


If ,y vu y

vu v v u

2= =

-l

l l

EXAMPLES

Differentiate

1. 5 23 5

xx+

-

Solution

=

y vu u x v x

u v

3 5 5 2

3 5

where and= = - = +

=l l

( )

( ) ( )

( )

( )

yv

u v v u

x

x x

xx x

x

5 2

3 5 2 5 3 5

5 215 6 15 25

5 231

2

2

2

2

=-

=+

+ - -

=+

+ - +

=+

ll l

2. 1

4 5 2x

x x3

3

-

- +

Solution

=

y vu u x x v x

v x

4 5 2 1

3

where and3 3

2 2

= = - + = -

=u x12 5-l l

( )

( ) ( ) ( )

( )

( )

yv

u v v u

x

x x x x x

xx x x x x x

xx x

1

12 5 1 3 4 5 2

112 12 5 5 12 15 6

110 18 5

2

3 2

2 3 2 3

3 2

5 2 3 5 3 2

3 2

3 2

=-

=-

- - - - +

=-

- - + - + -

=-

- +

ll l

ch8.indd 486 7/25/09 5:34:16 PM


8.10 Exercises

1. Differentiate

(a) 2 1

1x -

(b) 5

3x

x+

(c) 4x

x2

3

-

(d) 5 1

3xx

+

-

(e) 7x

x2

-

(f) 3

5 4xx+

+

(g) 2x x

x2 -

(h) 24

xx

-

+

(i) 4 32 7xx

-

+

(j) 3 1

5xx

+

+

(k) 3 7

1xx

2 -

+

(l) 2 3

2xx2

-

(m) 54

xx

2

2

-

+

(n) 4x

x3

+

(o) 3

2 1x

x x3

+

+ -

(p) 3 4

2 1x

x x2

+

- -

(q) 1x x

x x2

3

- -

+

(r) 2( )x

x

5

2

+1

(s) 5 1

(2 9)xx 3

+

-

(t) (7 2)

1xx

4+

-

(u) (2 5)

(3 4)

x

x3

5

-

+

(v) 1

3 1xx

+

+

(w) 2 3

1xx

-

-

(x) ( 9)

1xx

2

2

-

+

2. Find the gradient of the tangent to

the curve 3 1

2yxx

=+

at the point

1,21c m .

3. If ( )f xxx

2 14 5

=-

+ evaluate ( )f 2l .

4. Find any values of x for which the gradient of the tangent to the

curve 2 14 1yxx

=-

- is equal to .2-

5. Given ( )f xx

x3

2=

+ fi nd x if

( )f x61

=l .


to the curve 2

yxx

=+

at the

point 4,32c m .


to the curve 31y

xx2

=+

- at the

point where 2x = .

Angle Between 2 Curves

To measure the angle between two curves, measure the angle between the tangents to the curves at that point.

ch8.indd 487 7/10/09 12:03:30 PM


EXAMPLE

Find the acute angle formed at the intersection of the curves y x2= and ( 2) .y x 2= -

Solution

The curves intersect at the point (1,1) .

( , ), ( )

( )

( )

(1, 1), 2(1 2)

( )( )

tan

y x

dx

dyx

dx

dy

m

y x

dx

dyx

dx

dy

m

m m

m m

2

1 1 2 1

2

2

2 2

2

1

1 2 22 2

53 08

For

At

For

At

2

12

2

1 2

1 2

34

`

`

c

i

i

=

=

=

=

= -

= -

= -

= -

=+

-

=+ -

- -

=

= l

1

tanm m

m m

1 2

1 2i =

+

- where m1 and m2 are the gradients of the tangents to

the curves at the point of intersection.

ch8.indd 488 6/29/09 1:02:57 PM


1. (a) Sketch the curves 4y x2= - and 8 12y x x2= - + on the same set of axes.

Show that the curves intersect (b) at the point (2, 0) .Q

Find the gradient of the (c) tangent of each curve at point Q .

Find the acute angle at which (d) the curves intersect at Q .

2. (a) Sketch the curve y x2= and the line 6 9y x= - on the same set of axes.

Find the point (b) P , their point of intersection.

Find the gradient of the curve (c) y x2= at P .

Find the acute angle between (d) the curve and the line at P .

3. Find the acute angle between the curves y x2= and y x3= at point (1,1) .

4. Find the acute angle between the curves y x3= and 2 2y x x2= - + at their point of intersection.

5. What is the obtuse angle between the curves ( ) 4f x x x2= - and ( ) 12g x x2= - at the point where they meet?

6. The curves 2 4y x x2= - and 4y x x2= - + intersect at two points X and Y .

Find the coordinates of (a) X and Y .

Find the gradient of the (b) tangent to each curve at X and Y .

Find the acute angle between (c) the curves at X and Y .

7. Find the acute angle between the curve ( ) 1f x x2= - and the line ( ) 3 1g x x= - at their 2 points of intersection.

8. (a) Find the points of intersection between y x3= and 2y x x2= + .

Find the acute angle between (b) the curves at these points.

9. Show that the acute angle between the curves y x2= and 4y x x2= - is the same at both the points of intersection.

10. Find the obtuse angles between the curves 2y x x3= + and 5 2y x x2= - at their points of intersection.

8.11 Exercises

ch8.indd 489 7/10/09 12:16:51 PM


1. Sketch the derivative function of each graph

(a)

(b)

2. Differentiate 5 3 2y x x2= - + from fi rst principles .

3. Differentiate (a) 7 3 8 4x x x x6 3 2- + - - (b) 2 13 4

xx+

-

(c) ( 4 2)x x2 9+ - (d) ( )x x5 2 1 4- (e) x x2

(f) 5x2

4. Find dtdv if 2 3 4v t t2= - - .

5. Given ( ) (4 3) ,f x x 5= - fi nd the value of (a) (1)f (b) ( )f 11 .

6. Find the gradient of the tangent to the curve 3 5y x x x3 2= - + - at the point ( , ) .1 10- -

7. If 60 3 ,h t t2= - fi nd dtdh when .t 3=

8. Find all x -values that are not differentiable on the following curves .

(a)

(b)

(c)

9. Differentiate (a) f x x2 4 9 4= +] ]g g (b)

35y

x=

-

(c) 3 1y x x 2= -] g (d) 4y x=

(e) ( )f x x5=

Test Yourself 8

-5

2 3 4

21

345

-3-4

-2-1

y

x11-44 -3 -2 -1

-4

-5

-3 -2 -1 2 3 4

21

345

-3-4

-2-1 1

y

x

ch8.indd 490 6/29/09 1:03:16 PM


10. Sketch the derivative function of the following curve.

11. Find the equation of the tangent to the curve 5 3y x x2= + - at the point 2, 11 .^ h

12. Find the point on the curve 1y x x2= - + at which the tangent has a gradient of 3.

13. Find drdS if 4S r2r= .

14. At which points on the curve 2 9 60 3y x x x3 2= - - + are the tangents horizontal?

15. Find the equation of the tangent to the curve 2 5y x x2= + - that is parallel to the line 4 1.y x= -

16. Find the gradient of the tangent to the curve 3 1 2 1y x x3 2= - -] ]g g at the point where 2.x =

17. Find ( )f 4l when .f x x 3 9= -] ]g g

18. Find the equation of the tangent to the

curve 31yx

= at the point where .x61

=

19. Differentiate 21s ut at2= + with respect

to t and fi nd the value of t for which

5, 7dtds u= = and .a 10= -

20. Find the x -intercept of the tangent to

the curve 2 14 3y

xx

=+

- at the point where .x 1=

21. Find the acute angle between the curve y x2= and the line 2 3y x= + at each point of intersection.

22. Find the obtuse angle between the curve y x2= and the line 6 8y x= - at each point of intersection.

1. If ( ) 3 (1 2 ) ,f x x x2 5= - fi nd the value of (1)f and (1) .fl

2. If 7 15 3,Ahh

=-

+ fi nd dhdA when 1.h =

3. Given 2 100 ,x t t4 3= + fi nd dtdx and fi nd

values of t when 0.dtdx

=

4. Find the equations of the tangents to the curve ( 1)( 2)y x x x= - + at the points where the curve cuts the x -axis.

5. Find the points on the curve 6y x3= - where the tangents are parallel to the line 12 1.y x= - Hence fi nd the equations of the normals to the curve at those points.

Challenge Exercise 8

y

x

ch8.indd 491 6/29/09 1:03:48 PM


6. Find ( )f 2l if ( ) 3 2 .f x x= -

7. Differentiate (5 1) ( 9) .x x3 5+ -

8. Find the derivative of (4 9)

2 1 .yxx

4=

-

+

9. If ( ) 2 3 4,f x x x3 2= + + for what exact values of x is ( ) ?x 7=fl

10. Find the equation of the normal to the curve 3 1y x= + at the point where 8.x =

11. The tangent to the curve 2y ax3= + at the point where 3x = is inclined at 135c to the x -axis. Find the value of a .

12. The normal to the curve 1y x2= + at the point where 2,x = cuts the curve again at point P . Find the coordinates of P .

13. Find the exact values of the x - coordinates of the points on the curve (3 2 4)y x x2 3= - - where the tangent is horizontal.

14. Find the gradient of the normal to the curve 2 5y x x= - at the point (4, 8) .

15. Find the equation of the tangent to the curve 2 6y x x x3 2= - + + at point (1, 8) .P Find the coordinates of point Q where this tangent meets the y -axis and calculate the exact length of PQ .

16. (a) Show that the curves 3 2y x 5= -] g and

1

5 3yxx

=+

- intersect at 1, 1^ h Find the acute angle between the (b)

curves at this point .

17. The equation of the tangent to the curve 3 2y x nx x4 2= - + - at the point where 2x = - is given by 3 2 0.x y- - = Evaluate n .

18. The function ( ) 3 1f x x= + has a tangent that makes an angle of 30c with the x -axis. Find the coordinates of the point of contact for this tangent and fi nd its equation in exact form.

19. Find all x values of the function ( ) ( 3)(2 1)f x x x2 8= - - for which ( )f x 0=l .

20. (a) Find any points at which the graph below is not differentiable.

Sketch the derivative function for (b) the graph.

21. Find the point of intersection between the tangents to the curve 2 5 3y x x x3 2= - - + at the points where 2x = and 1.x = -

22. Find the equation of the tangent to the

parabola 2

3y x2

=- at the point where

the tangent is perpendicular to the line .x y3 3 0+ - =

23. Differentiate .x

x2

3 23

-

24. (a) Find the equations of the tangents to the parabola 2y x2= at the points where the line 6 8 1 0x y- + = intersects with the parabola.

Show that the tangents are (b) perpendicular.

y

x90c 180c 270c 360c

ch8.indd 492 6/29/09 1:04:22 PM


25. Find any x values of the function

( )8 12

2f xx x x3 2

=- +

where it is not

differentiable.

26. The equation of the tangent to the curve 7 6 9y x x x3 2= + - - is y ax b= + at the point where .x 4= - Evaluate a and b .

27. Find the exact gradient with rational denominator of the tangent to the curve 3y x2= - at the point where 5.x =

28. The tangent to the curve y xp

= has a

gradient of 61

- at the point where 3.x =

Evaluate p .

29. Find drdV when

32r r

= and 6h = given

31V r h3r= .

30. Evaluate k if the function ( ) 2 1f x x kx3 2= - + has ( ) .f 2 8=l

31. Find the equation of the chord joining the points of contact of the tangents to the curve 4y x x2= - - with gradients 3 and .1-

32. Find the equation of the straight line passing through 4, 3^ h and parallel to the tangent to the curve y x4= at the point 1, 1 .^ h

33. Find ( )7fl as a fraction, given

( )1

1f xx3

=+

.

34. For the function ( ) , ( ) ,f x ax bx c f 2 42= + + =

( ) ( )x1 0 8and= =f fl l when .x 3= - Evaluate a , b and c .

35. Find the equation of the tangent to the curve 2 2S r rh2r r= + at the point where 2r = ( h is a constant) .

36. Differentiate (a) 2 3 5x x x3 4- -] g (b) ( 3)

2 1xx

3-

+

37. The tangents to the curve 2 3y x x3 2= - + at points A and B are perpendicular to the tangent at , .2 3^ h Find the exact values of x at A and B .

38. (a) Find the equation of the normal to the curve 1y x x2= + - at the point P where 3.x =

Find the coordinates of (b) Q , the point where the normal intersects the parabola again.

ch8.indd 493 7/10/09 12:04:11 PM

Introduction to Calculusweb2.hunterspt-h.schools.nsw.edu.au/studentshared/MATHEMATICS… ·...

Documents

Transcript of Introduction to Calculusweb2.hunterspt-h.schools.nsw.edu.au/studentshared/MATHEMATICS… ·...