Differentiation Calculus was developed in the 17 th century by Sir Issac Newton and Gottfried...

DifferentiationCalculus was developed in the 17th century by Sir Issac Newton and Gottfried Leibniz who disagreed fiercely over who originated it.

Calculus provides a tool for solving problems involving motion. For example, methods obtained from calculus are used to study the orbits of planets, to calculate the flight path of a rocket and to predict the path of a particle through a magnetic field.

In fact calculus can be used to study any situation where a rate of increase or decrease is involved. (rate of change)

If we remember from standard grade distance time graphs. The speed was obtained from the gradient of the graph.

Time

Dis

tanc

e

DistanceSpeed =

Timed

tThis is the gradient of the line.

To determine the speed we needed two points on the graph.

This gave us the average speed between these two points.

But what about the speed at a certain time? The instantaneous speed.

x

y

x

h

x+h

f(x)

f(x+h)

change in

change in

ym

x

( ) ( )f x h f x

h

For an instantaneous gradient we require ‘h’ to be zero.

We label this gradient `( )f x

Let us examine this gradient as ‘h’ gets smaller and smaller.

In other words, let ‘h’ approach zero.

x

y

x

h

x+h

f(x)

f(x+h)

0

( ) ( )`( ) lim

h

f x h f xf x

h

`( ) is called the derived function or the derivative of .f x x

The derived function represents:

•The rate of change of the function•The gradient of the tangent to the graph of the function

Using the formula for f `(x) is called differentiating from first principles.

21. Derive the function ( )f x x

0

( ) ( )`( ) lim

h

f x h f xf x

h

2 2

0

( ) ( )limh

x h x

h

2 2 2

0

2limh

x xh h x

h

2

0

2limh

xh h

h

0

(2 )limh

h x h

h

0lim 2h

x h

2x

32. Derive the function ( )f x x

0

( ) ( )`( ) lim

h

f x h f xf x

h

3 3

0

( ) ( )limh

x h x

h

3 2 2 3 3

0

3 3limh

x x h xh h x

h

2 2 3

0

3 3limh

x h xh h

h

2 2

0

(3 3 )limh

h x xh h

h

2 2

0lim 3 3h

x xh h

23x

Numerous functions were derived in this way and a pattern was noticed.

( )

`( )

f x

f x

2 3 4 5 6x x x x x22 3x x 34x 45x 56x

In general terms:

1If ( ) , then `( ) , where is a rational number.n nf x ax f x nax n

8 3 4

1. Derive the following functions.

(a) ( ) ( ) 2 ( ) 5x b x c x d x

8( )f x x7`( ) 8f x x

1( )f x x0`( ) 1f x x

3( ) 2f x x2`( ) 6f x x

4( ) 5f x x5`( ) 20f x x

15

20

x

133

3


1 3(a) ( ) 3 ( ) 2 ( )

2x b x c x d

x1

31( )

2f x x

2

31`( )

6f x x

2

3

1

6x

3 2

1

6 x

( ) 3f x x1

23x1

23

`( )2

f x x

1

2

3

2x

3

2 x

3( ) 2f x x3

22x1

2`( ) 3f x x

3 x

1

3

3( )f x

x

1

33x

4

3`( )f x x

4

3

1

x

3 4

1

x

The derivative is also called the ‘rate of change’ or the ‘gradient of the tangent to the curve’. They all mean find the derived function.

1. Find the gradient of the tangent to the curve f(x) = x2 at x = 5.

2( )f x x

`( ) 2f x x

When 5, x `( ) 2 5 10f x

The gradient of the tangent at 5 is 10.x

2. Smoke form a factory chimney travels in t hours. Calculate the speed (rate of change) of the smoke after 4 hours.

t km

The distance travelled by the smoke is ( )d t t1

2t1

21

`( )2

d t t

1

2 t

After 4 hours, t = 4. 1

`(4)2 4

d 1

4

The speed of the smoke after 4 hours is ¼ km / h.

2 Exercise 6E

We can also derive functions with more than one term.

2 4


3(a) 2 3 ( ) ( ) 3 5

4

xx x b c x x

x

2 4( ) 2 3f x x x 3`( ) 4 12f x x x

1 11( ) 3

4f x x x

21`( ) 3

4f x x

2

1 3

4 x

11 0 2( ) 3 5f x x x x

1

21

`( ) 32

f x x

13

2 x

Derivatives of products and quotients We can also derive more complicated expressions. Before differentiating a function we have to express it as the sum of individual terms.

1. Derive ( ) ( 2)( 4)f x x x

2( ) 2 8f x x x

`( ) 2 2f x x

32. Derive ( ) ( )f x x x x 1 1

32 2( ) ( )f x x x x 7

2x x

5

27

`( ) 12

f x x 57

12

x

2 2 13. Derive ( )

x xf x

x

122( ) ( 2 1)f x x x x

3 1 1

2 2 22x x x

1 1 3

2 2 23 1

`( )2 2

f x x x x

3

3 1 1

2 2

x

x x

Now there will be times when you have a correct answer but it will not match the one shown in the back of the book. However they could well be the same. Look at our previous answer.

2

3 3

3 1 1 3 2 1

2 2 2

x x x

x x x

Always check for equality when checking your answers.

1. A designer of an artificial ski slope describes the shape of the slope by the function

3

24 2( )

d dh d

d

h(d)

d0

d is the horizontal distance in metres and h(d) is the height in metres. Calculate the gradient of the slope 4 metres horizontally from the start of the slope.

1

2( ) 4 2h d d d 1

2`( ) 2 2h d d

2

2d

2`(4) 2

4h 1

The gradient of the slope 4 metres after the start of the slope is -1.

Instead of writing the derivative as f `(x) we can use the Leibniz notation.

This is a geometric notation.

The Greek letter delta Δ means ‘change’. So going back to our gradient diagram,

x

y

Δx

Δy

0Gradient = lim

x

y

x

This is normally writtendy

dx

`( )dy

f xdx

1. Find the derivative of y = 3x4 -5x2 – 3x + 6

4 23 5 3 6y x x x

312 10 3dy

x xdx

2. Find the rate of change of 4 at 9.y t t 1

24y t1

22dy

tdt

At t = 9,2

9

dy

dt

2

t

2

3

1

23. Differentiate 8 with respect to .r p p p 1

28r p p 1

24 1dr

pdp

41

p

2

( 3)( 5)4. Find the gradient of the tangent to the curve at 5.

x xy x

x

21 2

2

2 151 2 15

x xy x x

x

2 32 30dy

x xdx

2 3

2 30

x x

2 30When 5,

25 125

dyx

dx

40 8

125 25

8The gradient of the tangent to the curve is

25

Equations of TangentsSince a tangent is a straight line, its equation may be given by

( )y b m x a

Hence to find the equation of a tangent we need to determine:

•the coordinates (a,b) of the point•the gradient, (rate of change) at that point.

31. Find the equation of the tangent to at 9.y x x

3when 9, 9x y 27 Hence (9,27) is the point.

Now we need the gradient at the point (9,27).

3

2y x1

23

2

dyx

dx

3

2

x

3 9When 9,

2

dyx

dx

9

2

Using ( )y b m x a 9

27 ( 9)2

y x

2 54 9 81y x 2 9 27y x 9 27

2 2y x

Increasing and decreasing functions

This function is increasing This function is decreasing

It has a positive gradient It has a negative gradient

If f(x) increases as x increases then the function is increasing.

If f(x) decreases as x increases then the function is decreasing.

Let us look at 3 points on the graph.

x

y

C

B

A

The function is decreasing between A and B

The function is increasing between B and C

What is the function doing at B?

If we look at the tangent to the curve at point ‘B’.

We see it is horizontal. It will therefore have zero gradient. It has no rate of change.

A to B B B to C

dy

dxnegative positivezero

The derivative of a function may be used to identify the shape of its curve.

21. Find the interval for which the function 3 2 5 is increasing.y x x

23 2 5y x x

6 2dy

xdx

The function is increasing when 0.dy

dx

6 2 0x 6 2x

1

3x

3 22. Find the intervals in which the function ( ) 3 8

is increasing and decreasing.

f x x x

3 2( ) 3 8f x x x 2`( ) 3 6f x x x

Let us first consider when the function is Increasing

23 6 0x x 3 ( 2) 0x x (this is a smile parabola with roots at x = 0 and x = 2)

0 2

When 0, 2 the gradient is positive so the function is increasing.x x

NB This is a graph of the GRADIENTnot the function.

When 0 2 the gradient is negative so the function is decreasing.x

3 23. Show algebraically that the function ( ) 3 3 10

is never decreasing.

f x x x x

3 2( ) 3 3 10f x x x x 2`( ) 3 6 3f x x x

23( 2 1)x x 23( 1)x

The value of (x – 1)2 must always be positive for all values of x.

This means that the value of 3(x – 1)2 must always be positive which by default means that it can never be negative for all values of x.

Since the gradient is never negative the function is never decreasing.

Stationary points

We saw previously that if f `(x) = 0 the function is neither increasing or decreasing.

The tangent is horizontal so the point (x, f(x)) is a stationary point.

Stationary points occur when f `(x) = 0.

The nature of the stationary point depends on the gradient either side of it.

Let us remind our self of what we found earlier.

x

y

C

B

AA to B B B to C

dy

dxnegative positivezero

x

y

Maximum turning point

x

y

Minimum turning point

x

y

Rising point of inflexion

x

y

falling point of inflexion

3 21. Find the stationary points on the curve ( ) 2 9 12 and

determine their nature.

f x x x x

3 2( ) 2 9 12f x x x x 2`( ) 6 18 12f x x x 0 for s.p.

26( 3 2) 0x x 6( 2)( 1) 0x x

2 and 1x x 3 2When 2, 2(2 ) 9(2 ) 12(2)x y 4 (2,4)

3 2When 1, 2(1 ) 9(1 ) 12(1)x y 5 (1,5)

We will now consider the gradient either side of the stationary points to determine their nature.

`( ) 6( 2)( 1)f x x x

6

( 2)

( 1)

`( )

x

x

f x

slope

x

2 and 1x x

1 1 1 2 2

+

-

-

+ 0

+

-

+

- 0

+

+

+

+

Maximum turning point at (1,5) and minimum turning point at (2,4).

3 42. Find the stationary points on the curve 4 and

determine their nature.

y x x

3 44y x x 2 312 4

dyx x

dx 0 for s.p.

24 (3 ) 0x x

0 and 3x x

3 4When 0, 4(0 ) (0 )x y 0 (0,0)3 4When 3, 4(3 ) (3 )x y 27 (3,27)

We will now consider the gradient either side of the stationary points to determine their nature.

24 (3 )dy

x xdx

24

(3 )

x

x

dy

dx

slope

x

0 and 3x x

0 0 0 3 3

+

+

+ 0

+

+

+ 0

+

-

-

Rising point of inflexion at (0,0) and maximum turning point at (3,27)

Curve sketchingWhen sketching curves, gather as much information as possible form the list below.

1. Intercepts.

(a) Where does the curve cut the y axis? x = 0(b) Where does the curve cut the x axis (roots)? y = 0

2. Stationary Points.

(a) Where is the curve stationary? f `(x) = 0(b) What are the nature of the stationary points?

3. Behaviour.

(a) What happens for large + / - values of x?

3 21. Sketch the curve 3y x x

when 0, 0.x y (0,0)

3 2when 0, 3 0y x x 2 ( 3) 0x x

0 and 3x x (0,0), (3,0)

3 23y x x

23 6dy

x xdx

0 for s.p.

3 ( 2) 0x x

0 and 2x x

3 2When 0, (0 ) 3(0 )x y 0 (0,0)

3 2When 2, (2 ) 3(2 )x y 4 (2, 4)

3 ( 2)dy

x xdx

3

( 2)

x

x

dy

dx

slope

x

0 and 2x x

0 0 0 2 2

-

-

+ 0

+

-

- 0

+

+

+

Maximum turning point (0,0) and minimum turning point at (2,-4)

3 23y x x

As ,x y As ,x y

Now lest put all this information onto a set of axis.

x

y

3

(2,-4)

0

3 23y x x

As ,x y As ,x y

Now lest put all this information onto a set of axis.

x

y

3

(2,-4)

0

3 23y x x

Closed IntervalsIn a closed interval the maximum and minimum values of a function are either at the stationary points or at the end points of the interval.

x

y

0

(-2,-12)

1

For the interval 2 1x

Maximum value = 0

Minimum value = -12


Maximum value = 18

Minimum value = -2

x

y

0

(-1,-2)

1

(3,18)

x

y

0

(-2,-12)

1

(4,48)


Maximum value = 48

Minimum value = -12

3 2

1. Find the maximum and minimum values of the function

( ) 4 4 1 in the closed interval 1 2f x x x x x

when 1, 4 1 4 1 0 ( 1,0) x y

when 2, 32 4 8 1 21 (2,21)x y

2`( ) 12 2 4 0 for s.p.f x x x 22(6 2) 0x x

2(3 2)(2 1) 0x x 2 1

and 3 2

x

2 32 4 8 25 2 25when , 1 ,

3 27 9 3 27 3 27x y

1 1 1 9 1 9when , 2 1 ,

2 2 4 4 2 4

x y


Maximum value = 21

Minimum value = -25/27

Graph of the derived functionWe can sketch the graph of the derived function f `(x) by looking at the features of the graph of f(x).

This is a simple strategy but does require an understanding that you are drawing the graph of f `(x) from the graph of f(x).

x

f(x)

( )y f x

x

f `(x)

x

f(x)

( )y f x

x

f `(x)

`( )y f x

x

f(x)

( )y f x

x

f `(x)

Draw the graph of the derived function.

x

f(x)

( )y f x


x

f `(x)

x

f(x)

( )y f x

x

f `(x)


x

f(x)

( )y f x

x

f `(x)


`( )y f x

Optimization: Maxima and minima

Without differentiation we would need to plot most functions to find their maximum and minimum values.

By finding stationary points we can easily identify and determine these values.

This allows us to examine all functions and find the functions maximum and / or minimum values. We can apply this to real life situations.

x

The glass front of a carriage clock has breadth x cm. The jeweller has used 16 cm of gold leaf to edge the perimeter of the glass.

(a) Find, in terms of x, expressions for the height of the glass front and the area of the glass.

(b) Find the dimensions of the rectangle that gives the maximum area.

(c) Calculate the maximum area.

( ) perimeter 2 2a h x 16 2 2h x 2 16 2h x

8h x

Area l b (8 ) x x

2( ) 8 A x x x

(b) Maximum and minimum area occurs when `( ) 0A x

2( ) 8 A x x x

`( ) 8 2 0 for s.p. A x x2 8x

4x We now need to determine whether thisgives a maximum or minimum area.

`( ) 8 2A x x

slope

x 4 4 4

+ -0

Hence maximum area occurs whenx = 4

The dimensions that give the maximum area are breadth = 4cm, height = 4cm.

( ) Maximum area = 4 4c 216cm

Further Applications

In many cases the solution to a problem cannot be found directly from the information provided.

Often an expression must be found for a value that is provided and this expression then used to obtain a function that can be differentiated.

A useful strategy:

Identify from the question the differential you require. For example

if it's a maximum or minimum volume it will be .

If it's an area then it will be etc.

dV

dxdA

dx

If the differential is , you need a formula for in terms of .dV

V xdx

If the differential is , you need a formula for A in terms of .dA

xdx

Use your algebra skills and the information in the question to

obtain the formula.

The volume of the square based glass display case shown is 500cm3. The length of the base is x cm. The base of the case is not made of glass.

x

x

(a) Show that the area of glass in the case is 2 2000

( )A x xx

(b) Find the dimensions of the case that minimise the use of glass and calculate this minimum area.

2( ) ( ) 4a A x x xh This formula has a ‘h’ in it that we do not need.

We need to find out what ‘h’ is in terms of x.

Is there any information in the question that will let us do this?

We are told that V = 500 so let us use that.

V l b h 2500 x h

2

500h

x Let’s substitute this into the area formula.

2( ) 4A x x xh 2

2

5004x x

x

2 2000x

x

(b) Max / Min area occurs when `( ) 0A x

2 1( ) 2000A x x x

2`( ) 2 2000A x x x 0 for s.p.

2

20002 0x

x

32 2000 0x 3 1000x

10x We will now determine the nature of the s.p.

`( )A x

slope

x 10 10 10

- +0

Hence minimum area occurs when x = 10

2`( ) 2 2000A x x x

500When 10, 5

100x h

Hence the dimensions of the case are length =10cm, breadth = 10cm and height = 5cm.

22000100 300

10Area cm

Differentiation Calculus was developed in the 17 th century by Sir Issac Newton and Gottfried...

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Transcript of Differentiation Calculus was developed in the 17 th century by Sir Issac Newton and Gottfried...