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TERMINOLOGY

8 Introduction toCalculus

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

Continuity: Describing a line or curve that is unbrokenover its domain

Continuous function: A function is continuous over aninterval if it has no break in its graph. For every x valueon the graph the limit exists and equals the functionvalue

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

Derivative function: The gradient function of a curveobtained through differentiation

Differentiable function: A function which is continuousand where the gradient exists at all points on thefunction

Differentiation: The process of finding the gradient of atangent to a curve which is called the derivative

Differentiation from first principles: The process of findingthe gradient of a tangent to a curve by finding thegradient of the secant between two points and findingthe limit as the secant becomes a tangent

Gradient of a secant: The gradient (slope) of the linebetween 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 thederivative of the function

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

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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, inflation, unemployment, filling 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 usedto 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 find 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

gures.

However, it wasn’t until the 17th 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 rst wrote about his calculus methods, which he called uxions, in 1671, but his Method

of 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 rst!

Search the Internet for further details on these two famous mathematicians. You can 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

74

75 77

76 78

79 81 83 85 87 89

80 82 84 86 88

91 93 95 97 99

90 92 94 96 98

01 03 05 07

00 02 04 06

T h o u s a n d B a r r e l s p e r D a y

January 1973–May 2007

You will learn about

integration in the

HSC Course.

Isaac Newton

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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.

runrise

m =

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 find 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 find 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.

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EXAMPLES

1. The graph shows the average distance travelled by a car over time.

Find the gradient and describe it as a rate.

Hours

k m

400

5 t

d

Solution

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

runrise

m

5400

1

80

80

=

=

=

=

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

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

several weeks.

ee s

N u m e r o

c a s e s

s )

1

N

Find the gradient and describe it as a rate.

CONTINUED

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Solution

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

m

10

1500

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 finding the gradient of a straight line in the number plane, we think ofa 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, traffic 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

k m

20

15

10

5

4321

t

Hours

k m

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

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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.

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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

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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 -2

x

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 nding the gradient of a straight line, we 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.

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Drawing tangents to a curve is difficult. 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

thex

-axis and wherem

is negative, the gradient function is below thex

-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

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To the left of ,x1

the gradient is negative, so this part of the graph will

be below the x -axis. Between x1 and ,x

2 the gradient is positive, so the

graph will be above the x -axis. Between x2 and ,x

3 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

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8.1 Exercises

Sketch the gradient function for each graph.

The gradient is zero at x1 and .x

2 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 ,x

2 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.

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7.

8.

9.

10.

Differentiation from First Principles

Seeing where the gradient of a curve is positive, negative or zero is a good rst step,

but there are methods to nd a formula for the gradient of a tangent to a curve.

The process of 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 tangentcan 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 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.

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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 derivativeexists 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 isdiscontinuous at this point.

c

y

x

The curve is not differentiable at

point c since it is not smooth at that

point.

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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 x x x

x x

1

3 2 1

for

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 finishes and the other starts, at f (1).

f

f

( )

( )

f x x

f x x

1 1

13 2

1 3 1 2

1

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

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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 = x 2

y = 3 x -2

y

x

1. y

x

2.

x 1

y

x

3.

x 1

y

x

4. y

x

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5.

x 1 x 2

y

x

6. ( )4

f x x=

7.3

1 y

x= -

+

8. ( )f x x x

x x

2

1 2

if

if

32

#=

+

)

9. ( )f x

x x

x

x x

2 3

3 2 3

1 2

for

for

for2

2

1

# #= -

- -

Z

[

\

]]

]

10.

-4

-4

-5

-3

-3

-2

-2

-1-1

2

1

3

4

5

1 2 3 4

x

y

11. tan y x= for x0 360c c# #

12. ( )f x xx

=

13. ( ) cosf 3 2i i= -

14. ( ) sin g 2z z=

15.9

3 y

x

x2

=

-

-

Limits

To differentiate from 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 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 x

x

x x

x2

2

2

1 2

2 1

3

x x

x

2

2

2

2

-

- -=

-

+ -

= +

= +

=

" "

"

You did this in

Chapter 5.

CONTINUED

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2. Find an expression in terms of x for2 3

limh

xh h h0

2

h

- -

"

.

Solution

( )

( )lim lim

limh

xh h h

h

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 for3 5

limx

x x x x0

2 2

x d

d d d+ -

"d .

Solution

( )

( )

lim lim

lim

x

x x x x

x

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

d

d d d

d

d d

d

+ -=

+ -

= + -

= -

" "

"

d d

d

1. Evaluate

(a)3

lim xx x

0

2

x

+

"

(b)5 2 7

lim xx x x

0

3 2

x

- -

"

(c)33

limx

x x3

2

x -

-

"

(d)416

limt

t 4

2

t -

-

"

(e)1

1lim

g

g

1

2

g -

-

"

(f)2

2lim

xx x

2

2

x +

+ -

" -

(g)

2

lim h

h h0

5

h

+

"

(h)3

7 12lim

xx x

3

2

x -

- +

"

(i)525

limn

n5

2

n -

-

"

(j)1

4 3lim

x

x x1 2

2

x-

+ +

" -

2. Find as an expression in terms of x

(a)2 4

limh

x h xh h0

2

h

- -

"

(b)2

limh

x h xh h0

3

h

+ -

"

(c)3 7 4

limh

x h xh h h0

2 2 2

h

- + -

"

(d)4 4

limh

x h x h xh0

4 2 2

h

- -

"

(e)3 4 3

limh

x h xh xh h0

2 2 2

h

+ - +

"

(f)2 5 6

limh

x h xh h0

2 2

h

+ +

"

(g)2

limx

x x x x

0

2 2

x d

d d-

"d

(h)4 2

limx

x x x0

2 2

x d

d d-

"d

(i)3

limx

x x x x x0

3 2

x d

d d d+ -

"d

(j)2 9

limx

x x x x x0

2

x d

d d d- +

"d

8.3 Exercises

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Differentiation as a limit

The formula m x x

y y

2 1

2 1=

-

-

is used to find 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 first principles , we first 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 magnified, 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

x f 1( x )= x 2

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

parabola?

f 1( x )= x 2

2.99

7.99 y

x

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

locally straight.

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Before using limits to find different formulae for differentiating from first

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 find 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 functionf x x

3=] g , find 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

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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

. , .f 2 1 2 1]_ gi.

( . ) ( )

..

.

mf f

2 1 2

2 1 2

2 1 22 1 2

12 61

3 3

=-

-

=-

-

=

. , .f 2 01 2 01]_ gi.

( . ) ( )

.

.

.

mf f

2 01 2

2 01 2

2 01 2

2 01 2

12 0601

3 3

=-

-

=

-

-

=

. , .f 2 001 2 001]_ gi.

( . ) ( )

.

.

.

mf f

2 001 2

2 001 2

2 001 2

2 001 2

12 006001

3 3

=-

-

=-

-

=

. , .f 1 9 1 9]_ gi.

( . ) ( )

.

.

.

mf f

1 9 2

1 9 2

1 9 2

1 9 2

11 41

3 3

=-

-

=-

-

=

. , .f 1 99 1 99]_ gi.

( . ) ( )

..

.

mf f

1 99 2

1 99 2

1 99 21 99 2

11 9401

3 3

=-

-

=-

-

=

. , .f 1 999 1 999]_ gi.

( . ) ( )

.

.

.

mf f

1 999 2

1 999 2

1 999 2

1 999 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 write2

( ) (2)12lim

x

f x f

2x -

-

="

.

Use y

m x x

y

2 1

12

-

-

= to find

the gradient of the secant.

CONTINUED

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2. For the curve y x2= , find 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 x

y y

f f

4 9 4 9

4 9 5

4 9 5

4 9 54 9 5

9 9

2 1

2 1

2 2

=

=-

-

=-

-

=-

-

=

^ h

(b) . , ( . )

.

( . ) ( )

..

.

B f

m x x

y y

f f

5 1 5 1

5 1 5

5 1 5

5 1 55 1 5

10 1

2 1

2 1

2 2

=

=-

-

=-

-

=-

-

=

^ h

(c) . , ( . )

.

( . ) ( )

..

.

B f

m x x

y y

f f

5 01 5 01

5 01 5

5 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 write5

( ) (5)10lim

x

f x f

5x -

-

="

.

We can find a general formula for differentiating from first 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 x

y y

x c

f x f c 2 1

2 1=

-

-

=-

-

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

this ( )c f l .

( )( ) ( )

limf c x c

f 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 x

y y

x h xf x h f x

h

f x h f x

2 1

2 1=

-

-

=+ -

+ -

=

+ -

To find the gradient of the secant, we make h smaller as shown, so that

Q becomes closer and closer to P .

P

Q

QQ

Q

y

x

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

( x, f ( x ))

Search the Internet using

keywords ‘differentiation from

first principles’, gradient of

secant’ and ‘tangent’ to find

mathematical websites that

show this working.

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

limh

f x h f x

0h

+ -

"

.

We call this ( )xf l .

( ) ( ) ( )limxh

f x h f xh 0

=+ -

"

f l

If we use , P x y ^ h and ,Q x x y y d d+ +^ h close to P where x y andd d are

small:

Gradient of secant PQ

m x x

y

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

y

x 0 d

d

"d . We

call thisdx

dy .

The symbol d is a

Greek letter called

delta.

limdx

dy

x

y

x 0 d

d=

"d

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

the tangent:

, ( ), ( ) , ( ),dx

dy

dx

d y

dx

d f x f x y l l^ h

These occur because Newton, Leibniz and other mathematicians over the

years have used different notation.

Investigation

Leibniz useddx

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.

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The three formulae for differentiating from first principles all work in a

similar way.

EXAMPLE

Differentiate from first principles to find the gradient of the tangent to

the curve 3 y x2= + at the point where 1.x =

Solution

Method 1:

( )( ) ( )

( )( ) ( )

( )( ) ( )

( )

( ) ( )

( )

lim

lim

lim

lim

lim

lim

lim

f c x c

f x f c

f x x

f

f c x c

f x f c

f x

f x f

x

x

x

x

x

x x

x

3

1 1 3

4

11

1

1

3 4

1

1

1

1 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 h

x

f h h

h h

h h

3

1 1 3

4

3

1

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 3 y x 2= -

is the same as ( ) .f x x 32= -

CONTINUED

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( )( ) ( )

( )( ) ( )

( )

( )

( )

lim

lim

lim

lim

lim

lim

f xh

f x h f x

f h

f h f

hh h

h

h h

h

h 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:

limdx

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 y 1 4d d+ + :

( )

( )

( )

lim

lim

y x

x x

x x

y x x

x

y

x

x 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

d

d d

d

d d

d

d

d

d

+ = + +

= + + +

= + +

= +

= +

=

+

= +

=

= +

= +

=

"

"

d

d

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

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EXAMPLE

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

Solution

f x x x

f x h x h x h

x xh h x h

x xh h x h

2 7 3

2 7 3

2 2 7 7 3

2 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 x

x xh h x h x x

xh h h

2 4 2 7 7 3 2 7 3

2 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

h

h 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 theother two formulae.

1. (a) Find the gradient of the secant

between the point ,1 2^ h and thepoint where 1.01x = , on the

curve 1. y x4= +

Find the gradient of the(b)

secant between ,1 2^ h and thepoint where 0.999x = on the

curve.

Use these results to find the(c)gradient of the tangent to the

curve 1 y x4= + at the point

,1 2^ h .

2. A function f x x x3= +] g has atangent 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

2

2

-

-

when 2.01x = .

Evaluate(c)2

( ) (2)

x

f x f

-

-

when

.x 1 99= .

Hence find the gradient of the(d)

tangent at the point ,2 10^ h .

3. For the function ,f x x 4

2= -

] g find the derivative at point P where 3x = by selecting points

near P and finding the gradient of

the secant.

4. If ( )f x x2= ,

find(a) ( )f x h+

show that(b)

( ) ( )f x h f x+ - xh h2 2= +

8.4 Exercises

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show that(c)( ) ( )

h

f x h f x+ - x h2= +

show that(d) ( )x x2=f l .

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 7

h

f x h f xx h

+ -

= + - .

Find(d) ( )xf l .

6. A function is given by

( ) 5f x x x2= + + .

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

( ) ( )5

h

f h f h

2 2+ -= + .

Find(e) (2)f l .

7. Given the curve ( ) 4 3f x x3= -

find(a) f 1-] g find(b) f h1 1- + -f -] ]g g find the gradient of the(c)

tangent to the curve at the point

where x 1= - .

8. For the parabola 1 y x2= -

find(a) f 3] g find(b) f h f 3 3+ -] ]g g find(c) (3)f l .

9. For the function

( )f x x x4 3 5 2= - - find(a) ( )f 1l

similarly, find the gradient(b)

of the tangent at the point

,2 10- -^ h .

10. For the parabola y x x22= +

show that(a)

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

by substituting the point

,x x y y d d+ +^ h

show that(b) 2 2x

y x x

d

dd= + +

find(c)dx

dy .

11. Differentiate from first principles

to find the gradient of the

tangent to the curve

(a) f x x2=] g at the point where1x =

(b) y x x2= + at the point ,2 6^ h (c) f x x2 52= -] g at the pointwhere x 3= -

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

where 2x =

(e) f x x x7 42= - -] g at thepoint ,1 6-^ h .

12. Find the derivative function for

each curve by differentiating

from first 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 1 y 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)

x

f x f

-

-

when

.x 3 999= . Evaluate(c)

4

( ) (4)

x

f 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

x x

11=

-

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evaluate(b)( ) ( )

x

f x f

5

5

-

-

when

5.01x = .

Use these results to nd the(c)

derivative of the function at the

point where x 5= .

15. Find the gradient of the tangent

to the curve4

y x2

= at point

, P 2 1^ h by nding the gradient ofthe 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

dy m= =

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

y

x y= k

So if , y kdx

dy 0= =

dx

d kx k=] g

dx

d k 0=] g

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Proof

Investigation

Differentiate from first principles:

y x

y x y x

2

3

4

=

=

=

Can you find a pattern? Could you predict what the result would be for xn ?

Alternatively, you could find an approximation to the derivative of a

function at any point by drawing the graph of.

( . ) ( ) y

f x f x

0 01

0 01=

+ -

.

Use a graphics calculator or graphing computer software to sketch the

derivative for these functions and find the equation of the derivative.

Mathematicians working with differentiation from first 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

dx

d

x nx

n 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 xx 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

h

h 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

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this proof.

EXAMPLE

Differentiate ( )f x x7= .

Solution

( ) 7f x x6=l

There are some more rules that give us short ways to differentiate functions.

The first one says that if there is a constant in front of the x (we call this a

coefficient), then it is just multiplied with the derivative.

dx

d

kx knx

n n 1=

-

^ h

( ) ( )dx

d kf x kf x= l^ h

A more general way of writing this rule is:

Proof

( )( ) ( )

[ ( ) ( )]

( ) ( )

( )

lim

lim

lim

dx

d 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 3x 8 .

Solution

3 y x

dx

dy x

x

3 8

24

If 8

7

7

#

=

=

=

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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:

( ) ( ) ( ) ( )dx

d f x g x f x g x+ = + ’l^ h

Proof

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

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

lim

lim

lim

lim

lim lim

dx

d 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

hf 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

You do not need to

know this proof.

EXAMPLE

Differentiate x x3 4+ .

Solution

( )dx

d x x x x3 43 4 2 3+ = +

EXAMPLES

Differentiate

1. 7x

Solution

dx

d x7 7=] g

CONTINUED

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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

dy x

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 5

6

4 2

4 2

= - +

- = - - - +

= -

l

l

5. Differentiate2

3 5x

x x2 +

Solution

Divide by 2x before differentiating.

x

x x

x

x

xx

x

dx

dy

23 5

23

25

23

25

2

3

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.

( )dr

dSr h

r h

2 2 2

4 2

r r

r r

= +

= +

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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 2

x

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= - + , finddxdy when

.x 2= -

6. Finddx

dy if

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

7. If s t t 5 202= - , finddt

ds .

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

9. Finddt

dv when v t 15 92= - .

10. If h t t 40 2 2= - , finddt dh .

11. Given34

V r 3r= , finddr

dV .

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

evaluate (1)f l .

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 whendx

dy 12=

15. Evaluate ( ) g 2l when

( ) g t t t t 3 4 2 13 2= - - + .

8.5 Exercises

Expand brackets

before differentiating.

Simplify by dividing


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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

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u EXAMPLES

1. Find the gradient of the tangent to the parabola y x 12= + at the

point ,1 2^ h .

Solution

, ( )

dx

dy x

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 curve2 6 1 y x x3 2= - + is equal to 18.

Solution

dx

dy x x6 122= -

dx

dy is the gradient of the tangent, so substitute

dx

dy 18= .

x 1= -

,

,

x x

x x

x x

x x

x x

x

18 6 12

0 6 12 18

2 3

3 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

,

dxdy x 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

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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 1

1 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

So1

#

=

= -

=

The normal is perpendicular to the tangent.

m m 1So1 2

= -

CONTINUED

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m

m

13 1

131

2

2

= -

= -

So the gradient of the normal is131

- .

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

dy x 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

So1 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- = -

- = - -

- = +

= - +

_

]]

i

g g


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 thepoint where x 1= -

(d) 5 2 3 y 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 pointwhere 3x =

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

where t 2=

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

point where 4r =

(i) 4h t t 4= - where t 0=

(j) f t t t t 3 8 55 3= - +] g at thepoint where 2t = .

8.6 Exercises

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2. Find the gradient of the normal

to the curve

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

(b) 3 5 2 y x x2= + - at the

point ,5 48-

^ h (c) f x x x2 72= - -] g at thepoint where 9x = -

(d) 3 2 y x x x3 2= + + - at the

point ,4 62- -^ h (e) f x x10=] g at the point where

1x = -

(f) 7 5 y x x2= + - at the

point ,7 5- -^ h (g) 2 3 1 A x x x3 2= + - + at the

point where 3x =

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

(i) 4 9V h h3= - + at the

point ,2 9^ h (j) 2 5 3 g x x x x4 2= - + -] g atthe point where x 1= - .

3. Find the gradient of the

(i) tangent and (ii) normal to

the curve

(a) 1 y x2= + at the point ,3 10^ h

(b) f x x5 2= -] g at the pointwhere x 4= -

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

where x 1= -

(d) 3 2 8 p 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 1 y x x4= - + at thepoint ,2 7^ h (b) ( ) 5 3 2 6f x x x x3 2= - - + at

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

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

where 2x =

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

where 2t =

5. Find the equation of the normal

to the curve

(a) f x x x3 53= - +] g at thepoint ,3 23^ h (b) 4 5 y x x2= - - at the point

,2 7-

^ h (c) f x x x7 2 2= -] g at the pointwhere 6x =

(d) 7 3 2 y x x2= - - at the point

,3 70-^ h (e) 2 4 1 y 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 8

2= - +

] g at thepoint ,1 11^ h (b) 2 5 y x x x3 2= + - at the

point ,3 6-^ h (c) 5 F x x x5 3= -] g at the pointwhere 1x =

(d) 8 7 y x x2= - + at the point

,3 8-^ h (e) 2 4 1 y x x x4 3= - + + at the

point where 1x = .

7.

For the curve 27 5, y x x

3= - -

find values of x for which 0dx

dy = .

8. Find the coordinates of the point

at which the curve 1 y 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 1 y 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 3 y x x2= - is parallel to the line

7 3 0x y - + = .

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12. Find the coordinates of point S

where the tangent to the curve

4 1 y x x2= + - is perpendicular

to the line 4 2 7 0x y + + = .

13. The curve 3 4 y 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 t 2= - + 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 xn works for any rational number n .

Investigation

(a) Show that1.1 1

( )x h x x x h

h

+- =

+

- .

Hence differentiate(b)1

y x= from first principles.

Differentiate(c) y x 1= - using a short method. Do you get the same

answer as 1(b)?

(a) Show that2. ( )( )x h x x h x h+ - + + = .

Hence differentiate(b) y x= from first principles.

Differentiate(c)2

y x=

1

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

EXAMPLES

1. Differentiate 7 x3 .

Solution

3

x x

dx

dy x

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.

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x

x

37 1

3

7

23

23

#=

=

2. Find the equation of the tangent to the curve4

y x2

= at the point

where 2.x =

Solution

y x

x

dx

dy x

x

4

4

8

8

2

2

3

3

=

=

= -

= -

-

-

2xWhen =

y 2

4

1

2=

=

Gradient of the tangent at 2, 1 :^ h

dx

dy

2

8

1

3= -

= -

Equation of the tangent:

y y m x x y x

x

y x

x y

1 1 2

2

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) 3x31

(g) 8x43

(h)2

2x-

-1

2. Find the derivative function,

writing the answer without

negative or fractional indices.

(a)1x

(b) 5 x

(c) x6

(d)2

x5

(e)5

x3-

(f)1

x

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(g)x2

16

(h) x x

(i)x3

2

(j)x x4

1 32 4

+


to the curve y x3= at the point

where 27.x =

4. If12

xt

= , nddt

dx when .t 2=

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

Evaluate .( )f 16l


to the curve2

3 y

x2= at the point

1, 121c m .

7. Finddx

dy if y x x

2= +^ h .

8. A function ( )2

f x x

= has a

tangent at , .4 1^ h Find thegradient of the tangent.9. Find the equation of the tangent

to the curve1

y x3

= at the point

2,81c m .


to ( ) 6f x x= at the point where

9.x =

11. (a) Differentiate xx

.

H(b) ence nd the gradient of thetangent to the curve y x

x= at

the point where 4.x =


to the curve4

y x= at the point

8,21c m .

13. If the gradient of the tangent to

y x= is6

1 at point A , nd the

coordinates of A .

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

( )f x43

=l . Evaluate x .

15. The hyperbola2

y x= has two

tangents with gradient252

- . Find

the coordinates of the points of

contact of these tangents.

Note that1

. x x 2 2

1 16 6

#=

Use index laws to

simplify rst.

Expand brackets rst.

Composite Function Rule

A composite function is a function composed of two or more other functions.

For example, 3 4x25

-^ 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

dx

du#=

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Proof

Let ,x y d 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

x

u

x u

x

y

u

y

x

u

Then

As

Sox u x0 0 0

" "

#

#

d

d

d

d

d

d

d d

d

d

d

d

d

d

=

=" " "d d d

Using the definition of the derivative from first principles, this gives

dx

dy

du

dy

dx

du#= .

You do not need to

learn this proof.

EXAMPLES

Differentiate

1. (5 4)x 7+

Solution

Let 5 4u x= +

` 7

( )

dx

du

y u

du

dy u

dxdy

dudy

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

dx

dux

y u

du

dy u

dx

dy

du

dy

dx

du

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

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3. 3 x-

Solution

2

2

2

2

2

( )

( 1)

(3 )

x xu x

dx

du

y u

du

dy u

dx

dy

du

dy

dx

du

u

x

x

3 33

1

21

21

21

2 3

1

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.

[dx

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

Proof

( )

( )

( )

( ) [ ( )]

u f x

dx

duf x

y u

du

dy nu

dx

dy

du

dy

dx

du

nu f x

f x n f x

Let

Then

n

n

n

n

1

1

1

`

#

#

=

=

=

=

=

=

=

-

-

-

l

l

l


this proof.

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EXAMPLES

Differentiate

1. (8 1)x3 5-

Solution

( ) [ ( )]

( )

( )

dx

dy f 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)

1

x 2+

Solution

( )

( )

( ) [ ( )]

( )

( )

( )

x

x

y f x n f x

x

x

x

6 1

16 1

6 2 6 1

12 6 1

6 1

12

n

2

2

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

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(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 21

x -

(r)( 1)

1

x2 4+

(s) ( )x7 3 2-3

(t)4

5

x+

(u)2 3 1

1

x -

(v)4(2 7)

3

x 9+

(w)3 3

1

x x x4 3- +

(x) ( )x4 1 43 +

(y)( )x7

154

-


to the curve 3 2 y x 3= -] g at the

point .,1 1^ h 3. If ( ) 2( 3)f x x2 5= - , evaluate (2)f l .

4. The curve 3 y x= - has atangent with gradient

21

at point

N . Find the coordinates of N .

5. For what values of x does the

function ( )4 1

1f x

x=

- have

( )f x494

= -l ?


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

dy u

dx

dv v

dx

du= +

Proof

y uv =

Given that , y u v andd 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

y u

x

v v

x

uu

x

v

since`

d d d

d d d d

d d d d d

d

d

d

d

d

dd

d

d

+ = + +

= + + +

= + + =

= + +

^ h

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0, 0

lim lim

lim lim lim

x u

x

y u

x

v v

x

uu

x

v

ux

v v

x

uu

x

v

dxdy u

dxdv v

dxdu

As

x x

x x x

0 0

0 0 0

" "d d

d

d

d

d

d

dd

d

d

d

d

d

dd

d

d

= + +

= + +

= +

" "

" " "

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.

You do not need to

know this proof.

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

dy x

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 x

x x

x

3 5 1 3 1

3 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

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.

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 g

g g

6 @

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

x

x

x

xx

x

x

x x x

x

x x

x

x x

x

x

3 5 2 221

5 2 3 4

3 5 2 3 4 5 2

3 5 2

5 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 2

15 6 3 4

5 2

19 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 3

x

x

-

+

8.9 Exercises

Change this into a product


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to the curve 2 3 2 y x x 4= -] g atthe point 1, 2^ h .

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

evaluate ( )1f l .

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 1 y x x2 4= -] g atthe point , .1 1

^ h


to ( 1) ( 1)h t t 2 7= + - at the point

, .2 9^ h 8. Find exact values of x for

which the gradient of the

tangent to the curve

2 3 y x x 2= +] g is 14.9. Given ( ) (4 1)(3 2)f x x x 2= - + ,

nd the equation of the

tangent at the point where

x 1= - .

Quotient Rule

Differentiating the quotient of two functions y v u

= gives the result.

dx

dy

v

v dx

duu

dx

dv

2=

-

Proof

y v u

=

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

`

( )

( )

( )

( )

( )

( ) ( )

( )

( )

( )

,

y y v v

u u

y v v

u uv u

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 v

vu v u uv u v

v v v

v u u v

x

y

v v v

v x

uu

x

v

x v 0 0

since

As " "

dd

d

dd

d

d

d

d

d

d

d d

d

d d

d

d d

d

d

d

d

d

d

d

d d

+ =

+

+

=+

+- =

=+

+

-+

+

=+

+ - +

=

+

+ - -

=+

-

=

+

-

a k

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( )lim lim

x

y

v v v

v x

uu

x

v

dx

dy

v

v dx

duu

dx

dv

x x0 0

2

d

d

d

d

d

d

d

=

+

-

=

-

" "d d

R

T

SSSS

V

X

WWWW

It is easier to remember this rule as y v

u v v u2

= -

l l l

.


this proof.

If , y v u

y v

u v v u2

= = -

l l l

EXAMPLES

Differentiate

1.

5 2

3 5

x

x

+

-

Solution

=

y v u

u x v x

u v

3 5 5 2

3 5

where and= = - = +

=l l

( )( ) ( )

( )

( )

y v

u v v u

xx x

x

x x

x

5 23 5 2 5 3 5

5 2

15 6 15 25

5 2

31

2

2

2

2

= -

=

+

+ - -

=

+

+ - +

=

+

l l l

2.

1

4 5 2

x

x x3

3

-

- +

Solution

=

y v u u x x v x

v x

4 5 2 1

3

where and3 3

2 2

= = - + = -

=u x12 5-l l

( )

( ) ( ) ( )

( )

( )

y v

u v v u

x

x x x x x

x

x x x x x x

x

x x

1

12 5 1 3 4 5 2

1

12 12 5 5 12 15 6

1

10 18 5

2

3 2

2 3 2 3

3 2

5 2 3 5 3 2

3 2

3 2

= -

=

-

- - - - +

=

-

- - + - + -

=

-

- +

l l l

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8.10 Exercises

1. Differentiate

(a)

2 1

1

x -

(b)5

3x

x

+

(c)4x

x2

3

-

(d)5 1

3

x

x

+

-

(e) 7

x

x2

-

(f)3

5 4xx

+

+

(g)2x x

x2

-

(h)24

x

x

-

+

(i)

4 32 7

x

x

-

+

(j)3 1

5x

x

+

+

(k)3 7

1

x

x2

-

+

(l)2 3

2x

x2

-

(m)54

xx2

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)

x

x 3

+

-

(t)(7 2)

1xx 4

+

-

(u)

(2 5)

(3 4)

x

x3

5

-

+

(v)

1

3 1

x

x

+

+

(w)2 3

1

x

x

-

-

(x)

( 9)

1

x

x2

2

-

+

2. Find the gradient of the tangent tothe curve

3 1

2 y

x

x=

+ at the point

1,21c m .

3. If ( )f xxx

2 1

4 5=

-

+ evaluate ( )f 2l .

4. Find any values of x for which

the gradient of the tangent to the

curve2 14 1

y x

x=

-

- is equal to .2-

5. Given ( )f xx

x3

2=+

nd x if

( )f x61

=l .


to the curve2

y x

x=

+ at the

point 4,32c m .


to the curve3

1 y

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.

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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

dy x

dx

dy

m

y x

dx

dy x

dx

dy

m

m m

m m

2

1 1 2 1

2

2

2 2

2

1

1 2 2

2 2

53 08

For

At

For

At

2

1

2

2

1 2

1 2

34

`

`

c

i

i

=

=

=

=

= -

= -

= -

= -

=+

-

=+ -

- -

=

= l

1tan

m m

m m

1 2

1 2i =

+

-

where m1 and m

2 are the gradients of the tangents to

the curves at the point of intersection.

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1. (a) Sketch the curves 4 y x2= -

and 8 12 y 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 9 y x= - on the same

set of axes.

Find the point(b) P , their pointof 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).


curves y x3= and 2 2 y x x2= - + at their point of intersection.

5. What is the obtuse angle between

the curves ( ) 4f x x x2= - and

( ) 12 g x x2= - at the point where

they meet?

6. The curves 2 4 y x x2= - and

4 y 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 .


curve ( ) 1f x x2= - and the line

( ) 3 1 g x x= - at their 2 points of

intersection.

8. (a) Find the points of intersection

between y x3= and 2 y 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

4 y x x2= - is the same at both

the points of intersection.

10. Find the obtuse angles betweenthe curves 2 y x x3= + and

5 2 y x x2= - at their points of

intersection.

8.11 Exercises

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1. Sketch the derivative function of

each graph

(a)

(b)

2. Differentiate 5 3 2 y x x2= - + from first

principles.

3. Differentiate

(a) 7 3 8 4x x x x6 3 2- + - -

(b)

2 1

3 4

x

x

+

-

(c) ( 4 2)x x2 9+ -

(d) ( )x x5 2 1 4-

(e) x x2

(f)5

x2

4. Finddt

dv if 2 3 4v t t 2= - - .

5. Given ( ) (4 3) ,f x x 5= - find the value of

(a) (1)f

(b) ( )f 11 .

6. Find the gradient of the tangent to the

curve 3 5 y x x x3 2= - + - at the point

( , ) .1 10- -

7. If 60 3 ,h t t 2= - finddt

dh 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)3

5 y

x=

-

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

(d)4

y x= (e) ( )f x x5=

Test Yourself 8

-5

2 4

21

4

5

-

-4

-2

-1

y

x

- - -2 -1

-4

5

-2 -1 2 4

2

1

4

5

4

2

1 1

y

x

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10. Sketch the derivative function of the

following curve.

11. Find the equation of the tangent to

the curve 5 3 y x x2= + - at the point

2, 11 .^ h

12. Find the point on the curve

1 y x x2= - + at which the tangent has a

gradient of 3.

13. Finddr

dS if 4S r 2r= .

14. At which points on the curve

2 9 60 3 y x x x3 2= - - + are the tangents

horizontal?

15. Find the equation of the tangent to the

curve 2 5 y x x2

= + - that is parallel tothe line 4 1. y x= -

16. Find the gradient of the tangent to the

curve 3 1 2 1 y x x3 2= - -] ]g g at the point

where 2.x =

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


curve31

y x

= at the point where .x61

=

19. Differentiate21

s ut at 2= + with respect

to t and find the value of t for which

5, 7dt

dsu= = and .a 10= -

20. Find the x -intercept of the tangent to

the curve2 14 3

y x

x=

+

- at the point where

.x 1=

21. Find the acute angle between the curve

y x2

= and the line 2 3 y x= + at eachpoint of intersection.

22. Find the obtuse angle between the curve

y x2= and the line 6 8 y x= - at each

point of intersection.

1. If ( ) 3 (1 2 ) ,f x x x2 5= - find the value of

(1)f and (1).f l

2. If7 1

5 3, A

h

h=

-

+ find

dh

dA when 1.h =

3. Given 2 100 ,x t t 4 3= + finddt

dx and find

values of t when 0.dt

dx=

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 6 y x3= -

where the tangents are parallel to the line

12 1. y x= - Hence find the equations of

the normals to the curve at those points.

Challenge Exercise 8

y

x

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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. y

x

x4

=

-

+

9. If ( ) 2 3 4,f x x x3 2= + + for what exact

values of x is ( ) ?x 7=f l

10. Find the equation of the normal to the

curve 3 1 y x= + at the point where

8.x =

11. The tangent to the curve 2 y 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 1 y 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 5 y x x= - at the point (4,8).

15. Find the equation of the tangent to

the curve 2 6 y 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 2 y x 5= -] g and

15 3 y xx=

+

- 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 2 y 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 withthe x -axis. Find the coordinates of the

point of contact for this tangent and find

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 3 y x x x3 2= - - + at the points where

2x = and 1.x = -


parabola2

3 y

x2=

- at the point where

the tangent is perpendicular to the line

.x y 3 3 0+ - =

23. Differentiate .x

x2

3 23

-

24. (a) Find the equations of the tangents

to the parabola 2 y x2= at the points

where the line 6 8 1 0x y - + = intersects

with the parabola.

Show that the tangents are(b)

perpendicular.

y

x

90c 180c 270c 360c

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25. Find any x values of the function

( ) 8 122f x x x x3 2=

- + where it is not

differentiable.

26. The equation of the tangent to the curve

7 6 9 y 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

3 y x2= - at the point where 5.x =

28.The tangent to the curve y x

p=

has agradient of

61

- at the point where 3.x =

Evaluate p .

29. Finddr

dV when

3

2r

r= and 6h = given

31

V 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 joiningthe points of contact of t

ch8.pdf

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