Example 1

3 2( ) 4 2 3 6f x x x x

3 2( ) 4 2 3 6f x x x x Find the derivative of

Function is written in terms of f(x), answer should be written in terms of f ′ (x)

2( ) 12 4 3f x x x

Continue

Example 2Find the derivative of

3 4

4

34

3

1

1

yx

y y x

x

3 4

1y

x

Function is written in terms of y = answer should

be written in terms of dy/dx =

7

3

3 7

4

34

3

dyx

dxdy

dx x

Continue

Example 3

2

( 3)( 2)

6

s t t

s t t

Differentiate s with respect to t for

( 3)( 2)s t t

Function is written in terms of s and t so

derivative will be ds/dt =

2 1ds

tdt

Continue

Example 4 Harder examples

11

2

11 2

4 3( )

3( ) 4

( ) 4 3

f xx x

f x x

x

f x x x

Find the derivative of 4 3

( )f xx x

32 2

32 2 32

3( ) 4

24 3 4 3

( ) ( )22

f x x x

f x f xx x xx

Continue

Example 5 Harder examples

Find the derivative of

22

y xx

2

22

2 2

2 2 2

44

4 4

y x y x xx x x

y xx

y x x

3

3

8 2

82

dyx x

dxdy

xdx x

Continue

Example 6 Harder examples

Find the derivative of

23

( ) 3f x xx

2

2

12 1 2 12

1

2

3 3 3( ) 3 ( ) 3 3

18 9( ) 9

18( ) 9 9 ( ) 9 18 9

f x x f x x xx x x

xf x x

xx

xf x x x f x x x x

x

122

2

18( ) 18 9

29 9

( ) 18

f x x x x

f x xxx

Example 7 Harder examples

Find the derivative of 2 5x

yx

2

2

1

5

5

5

xy

x

xy

x x

y x x

2

2

1 5

51

dyx

dxdy

dx x

Example 8 Harder examples

Find the derivative of 2 3

( )x

f xx

1 1

2 2

1 1

2 2

2 3( )

2 3( )

( ) 2 3

xf x

xx

f x

x x

f x x x

1 3

2 2

1 3

2 2

3

3( )

21 3

( )

21 3

( )2

f x x x

f x

x x

f xx x

Example 9 Harder examples

3 2

3 2

6 4

6 4

dyx x

dxdy

dx x x

Find the derivative of 2

(1 )(3 )x xy

x

2

2 2

2 2 2 2

2 1

(1 )(3 )

3 4 3 4

3 4 1

x xy

x

x x x xy y

x x x x

y x x

Stationary Points Example

A function is given by f (x) = 4x3 – x4. Find: (i) the coordinates and nature of the stationary points on the curve;

(ii) the intervals on which f (x) is increasing or decreasing.

Stationary Points

3 4(0) 4(0) (0)

(0) 0

f

f

3 4

2 3

( ) 4

( ) 12 4

f x x x

f x x x

Step 1 Find the stationary points

Stationary points occur when f ′ (x) = 02 3

2

2

12 4 0

4 (3 ) 0

4 0 or 3 0

0, 3

x x

x x

x x

x x

Finding y coordinates of function f (x)When x = 0,

When x = 3,

Stationary points occur at ( 0 , 0 ) and ( 3 , 27 )

3 4(0) 4(3) (3)

(0) 27

f

f

Continue

x 0 3

f ′(x)

slope

2 3(1) 12(1) 4(1)

(1) 8

f

f

Step 2 Determine the nature of the stationary points

Consider the gradient f ′ (x) around each stationary point:

2 3( 1) 12( 1) 4( 1)

( 1) 16

f

f

means x approaching 0, so test the gradient by picking a suitable value for x < 0 and substituting in to f ′(x)

pick a value for x between 0 and 3

pick a value for x > 3

2 3( ) 12 4f x x x

+ ve 0 + ve 0 − ve

2 3(10) 12(10) 4(10)

(10) 2800

f

f

At the point (0,0) the shape of the graph is + ve 0 + ve

∴ at the point (0,0) there is a rising point of inflexion

At the point (3,27) the shape of the graph is + ve 0 - ve ∴ at the point (3,27) there is a maximum turning point

Continue

(ii) From the table we can see the shape of the whole graph:

This tells us that f is increasing for values of x up to 3, except at 0.f is increasing for

and that f is decreasing for values of x greater than 3.

f is decreasing for

Stationary Points

x 0 3

f ′(x) + ve 0 + ve 0 − ve

slope ↗ → ↗ → ↘

> 3; x x R

< 3, 0; x x x R

BackBack

The curve cuts the x-axis when y = 0

the x-intercepts are at ( 0 , 0 ) and (3,0).

Curve Sketching

3 2

2

3 0

( 3) 0

0 or 3

x x

x x

x x

Sketch the curve y = x3 – 3x2.

Step 1 The y-interceptThe curve cuts the y-axis when x = 0

the y-intercept is at ( 0 , 0 ).

Step 2 The x-intercept

3 2(0) 3(0)

0

y

y

Continue

Curve Sketching

3 2

2

3

3 6

y x x

dyx x

dx

2

0

3 6 0

3 ( 2) 0

3 0 or 2 0

0 or 2

dy

dx

x x

x x

x x

x x

3 2(0) 3(0)

0

y

y

Step 3 Stationary points

For stationary values,

So

To find y coordinates when x = 0

3 2(2) 3(2)

4

y

y

To find x coordinates when x = 2

Stationary values occur at ( 0 , 0 ) and ( 2 , −4 )

Continue

Curve Sketching

x 0 2

slope

23 6dy

x xdx

dy

dx

Step 3 Stationary points cont’d.

The nature of the stationary values is best described through the gradient table as before

Stationary values occur at ( 0 , 0 ) and ( 2 , −4 )

Remember to use dy/dx and select suitable values to test the gradient.

(−1) = 9 dy

dx

+ ve 0 − ve 0 + ve

(1) = -3 (3) = 9

Maximum Turning Point at ( 0 , 0) and Minimum turning point at ( 2 , −4 )

Continue

Curve SketchingStep 4 Large positive and negative

values of x

The function y = x3 – 3x2 will behave like x3 for very large values of x.

If x is very large and positive

x3 is large and positive so y is large and positive

If x is very large and negative

x3 is large and negative so y is large and negative

Continue

Curve Sketchingy

x

Step 5 Sketch the graph

(0,0)

(2,-4)

(3,0)

y = x3 – 3x2

Closed IntervalsExample 1

y

x62

(2, −2)

(4, −5)

For the interval 2 ≤ x ≤ 6 sometimes written as [2,6]

The minimum value is −5 at x = 4

Min

Max

The maximum value is 0 at x = 6

Continue

Closed Intervals

3 2(1) 4(1) (1) 4(1) 1

(1) 0

f

f

3 2( 1) 4( 1) ( 1) 4( 1) 1

( 1) 0

f

f

Example 2Find the maximum and minimum values off(x) = 4x3 −x2 −4x + 1 in the closed intervals(a) −1 ≤ x ≤ 2 (b) [−1,1]

Maximum and minimum values are either at a stationary point or at an end point, so start by finding these points.

Step 1 End Points

3 2(2) 4(2) (2) 4(2) 1

(2) 21

f

f

Continue

Closed Intervals3 2

2

( ) 4 4 1

( ) 12 2 4

f x x x x

f x x x

( ) 0f x

2

( ) 0

12 2 4 0

2(2 1)(3 2) 0

2 1 0 or 3 2 0

1 2 or

2 3

f x

x x

x x

x x

x x

Step 2 Stationary Points

For stationary points

To find y-coordinates3 2

2 2 2 24 4 1

3 3 3 3

2 32 4 81

3 27 9 3

2 32 12 72 27

3 27 27 27 27

2 25

3 27

f

f

f

f

Stationary points occur at and 1 9

,2 4

2 25,

3 27

3 21 1 1 1

4 4 12 2 2 2

1 4 12 1

2 8 4

1 2 1 8 4

2 4 4 4 4

1 9

2 4

f

f

f

f

Continue

Closed Intervals

x

f ′ (x) + ve 0 − ve 0 + ve

slope ↗ → ↘ → ↗

2

3

1

2

Step 2 Stationary Points (Nature Table)

Maximum turning point at 1 9

,2 4

2 25,

3 27 And, minimum turning point at

Continue

21

-25/27

-1 2

f(x)

x

Closed IntervalsStep 3 Sketch the graph

(a) For -1 ≤ x ≤ 2

The maximum value of f is 21

The minimum value of f is -25/27

(b) For [-1,1] or -1 ≤ x ≤ 1

The maximum value of f is 9/4

The minimum value of f is -25/27

9/4

-25/27

-1 1

BackBack

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

(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.

x cm

x cm

This problem can best be solved using differentiation.

Find a formula for the volume in terms of x and h (height). Rearrange to get an equation for h in terms of x

Use that to find a formula for the total area of glass in terms of x.

h

Continue

Problem Solving

2

2

500

500

500

x x h

x h

hx

(a) The total area of glass is given by,

A(x) = area of top + area of 4 sides

To find the area of glass we need to find an expression for the height of the case.

Volume = length x breadth x height

x cmx cm

h

Total area of glass, A(x) = area of top + 4 x area of a side2

2

22

2

( ) 4( )

( ) 4

500( ) 4

2000( )

A x x x h

A x x xh

A x x xx

A x xx

Continue

Problem Solving

2

2 1

2000( )

( ) 2000

A x xx

A x x x

( ) 0A x

2

2

( ) 2 2000

2000( ) 2

A x x x

A x xx

(b) By finding the minimum stationary point for A(x) we can obtain the value of x for which the area is a minimum.

For a stationary point

2

3

3

3

3

20002 0

2 2000 0

2( 1000) 0

1000

1000

10

xx

x

x

x

x

x

2 2000(10) 10

10(10) 300

A

A

x 10

f ′ (x) - ve 0 + ve

slope ↘ → ↗

Check that the stationary point at 10 is a minimum

A minimum occurs when x = 10.When x = 10, length =10, breadth = 10 and height = 5

The minimum area of glass is

2

from

500h

x

Continue

Rate of ChangeThe number of bacteria, N(t), in a certain culture is calculated using the formula N(t) = 4t3 + 400t + 1000 where t is the time in hours from the start of the growth. Calculate the growth rate of this culture when t=5.

The rate of change is a derative so we need to find N ′(t).

3

2

2

( ) 4 400 1000

( ) 12 400

(5) 12(5) 400

(5) 700

N t t t

N t t

N

N

The growth rate of the culture is 700 bacteria per hour after 5 hours.

t=5, so substitute 5 into N’(t) for t.

BackBack

Velocity, Speed and Acceleration

The speed of a particle, relative to its starting position, is given by

3 2( ) 6 8 20d t t t t

where t is the time in seconds and d(t) is the distance in cm.

(a) Calculate the speed of the particle when t = 5

(b) (i) Find a function to describe the acceleration of the particle (ii) What is the acceleration when t = 5?

Continue

Velocity, Speed and Acceleration

The speed of the particle is its rate of change of distance with time, given by d’(t). So we need to differentiate :

3 2

2

2

( ) 6 8 20

( ) 3 12 8

at t = 5 seconds

(5) 3(5) 12(5) 8

(5) 143

d t t t t

d t t t

d

d

The speed of the particle after 5 seconds is143 cm/s

(a)

Continue

Velocity, Speed and Acceleration

(b) (i) To find the acceleration of the particle we need to find how quickly it is changing speed i.e. the rate of change of speed with time.

We find this by differentiating the speed.

The speed was found in (a) as d′(t). We can call this s(t) where s is speed.

Acceleration will then be s′(t)

3 2

2

( ) 6 8 20

( ) 3 12 8

( ) 6 12

( ) 6 12, when t = 5

(5) 6(5) 12

(5) 42

d t t t t

d t t t

s t t

s t t

s

s

(b) (ii)

So the acceleration of the particle at t = 5 is 42 cm/s per second