All Differentiation

Higher Unit 1

Differentiation

Introduction to differentiation

1.

2. The graph doesn’t realistically depict changes in speed

Introduction to Differential Calculus

Up to this moment, your experience of speed and the relationshipwith distance and time has been restricted to questions such as thoserepresented by this travel graph. However, there are two problems:1. We can only find the AVERAGE speed between 2 time points, we cannot find the actual speed when t = a given time.

Do not go from speed x to zeroinstantaneously.

0 1 2 3

50

40

30

20

10

“If I have seen further it is by standing on the shoulders of giants.”

Sir Isaac Newton(1642-1727)

Therefore we will look at methods for finding the instantaneousSpeed or rate of change for a given graph or function.

This is known as Differential Calculus and was developed by Sir Isaac Newton and others and is used in many areas of maths.

To Speed or Not To Speed?

A couple of motorists have been photographed by a speed camera and charged with speeding. The speed limit is under 30m/s in the area they were driving.

They claim they were under the speed limit.

Use the table below to determine if they broke the law or not. The camera flashed at 3 seconds.

Time 0s 1s 2s 3s 4s

Distance 0m 5m 20m 45m 80m

Finding The Average Speed.Time 0s 1s 2s 3s 4s

Distance 0m 5m 20m 45m 80m

The formula for finding the speed is:

Time

DistanceSpeed

Calculate the average speed between 0 and 3 seconds.

03

045

Speed = 15m/sSo far so good for our dastardly duo.

So far we have worked out the average speed between 0 and 3 seconds as shown below:

0 1 2 3

50

40

30

20

10

45m

3s

Now repeat for 1 to 3 seconds.

And 2 to 3 seconds

f=at2

The Instantaneous Speed.We use these results to complete the table below:

Time Interval 0-3 s 1-3 s 2-3 s

Average Speed 15m/s 20m/s 25m/s

It is clear that the closer the times are taken together the more accurately the speed of the car can be measured at 3 seconds.

How can we get more accurate readings ?

Use d = 5t 2 you snickering fools !!

Using the formula d = 5 t 2 we find the table below:

Time 2.6s 2.7s 2.8s 2.9s 3.0s

Distance.

Time Interval

2.6-3.0 2.7-3.0 2.8-3.0 2.9-3.0

Average Speed.

33.8m 36.45m 39.2m 42.04m 45m

28m/s 28.5m/s 29m/s 29.6m/s

Guilty or not guilty ? The verdict is yours !!!

The Tangent To The Curve.To calculate the average speed we have been calculating the gradients of lines as shown below:

0 1 2 3

50

40

30

20

10

To get the instantaneous speed of the car at three seconds we require to calculate the gradient of the line which is tangential to the curve at t = 3seconds.That is to say the line contacts the curve at almost a single point at t= 3. The line is shown below:

0 1 2 3

50

40

30

20

10

The tangent to the curve at t = 3 .

The Final Verdict.

To calculate the value of the gradient of the tangent you require to consider a time as close to 3 seconds as possible. You might consider t= 3 and t= 2.99s or t= 3 and t=2.999s and so on…

Investigate and find out if the car travelled at 30m/s.

Learn your calculus numb skulls or you’ll

never catch us !!

The Derived Function2.

Required skills

Before we start ……….

You will need to remember work with Indices, as well as what you have learned about Straight Lines from Unit 1.1.

Lets recall the rules on indices …..

Rule Examples

a0 = 1 12.3140 =

a-m = x -5 =

=

am × an = am+n 2a3/2 × 3a1/2 = =

am an = am-n x2 x -3 = =

(am)n = amn (q2)3= =

Rules of indices

3n

6q

1

am

1

n3

a amnm

n x23 y6

5

1

65 y

5

1

x23x

2 3x

3 12 26a a

26a

5x2 2 2q q q

What is Differentiation?

Differentiation is the process of deriving f ′(x) from f(x). We will look at this process in a

second. f ′ (x) is called the derived function or

derivative of f(x). The derived function represents:• the rate of change of the function• the gradient of the tangent to the graph of the

function.

Tangents to curvesThe derivative function is a measure of the gradient or slope of a function at any given point. This requires us to consider the gradient of a line.

We can do this if we think about how we measure the gradient from Unit 1.1

y2 – y1

x2 – x1

The gradient of AB = mAB

Change in y

Change in x

2 1

2 1x

y y

x2 1

2 1

, Gradient Formulax

y y

mx

A(x1,y1)

B(x2,y2)

C

Tangents to curves

We will look at a function and think about the gradient of the function at any given point. The function itself is not important, the process we go through to get the gradient is. We want to find the gradient of a curve.

A is the point (x, f(x)) and Bis a point on the function a short distance h from A.This gives B the coordinates(x+h, f(x+h))

The line AB is shown on the diagram. We want to find the gradient of the curve at A. If we find the gradient of the line AB and move B towards A we should get the gradient at A.

y

x

y = f(x)

x x h

A

B

2 1

2 1AB

y yM

x x

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

(( ), ( ))x f x

ABM

( )f x h

( )f x h

( )f x h

( )f x h

( )f x h( )f x h

( )f x

( )f x( )f x ( )f x

( )f x ( )f x ( )f x ( )f x( )f x

( )x h

( )x h

( )x h

( )x h

( )x h

( )x h

( )x

( )x

( )x( )x

( )x( )x ( )x ( )x ( )x ( )x ( )x

( )x

Gradient of a function

( ) ( )

( )AB

f x h f xM

x h x

What happens if we make smaller?h

y

x

y = f(x)

x x h

A

B

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

(( ), ( ))x f x

x hx hx hx hx hx hx hx hx hx hx hx h x hx h x

When

Then gradient of AB is the gradient of the tangent at B

x h x

B

y

x

y = f(x)

The gradient doesn't change as the line gets smaller and so the gradient at B must be the same as the gradient of the line

Shrinking the tangent line is the same as letting get very smallh

What are we saying?

( ) ( )

( )AB

f x h f xM

x h x

But

0

( ) ( )lim

( )B h

f x h f xM

x h x

We know

This is the basic, first principle definition of the derived function. Usually written / ( )f x

Putting it together….Differentiate the function f(x) = x2

from first principles.

The derivative is the same as the gradient of the tangent to the curve so we can go straight to the gradient formula we saw in the previous slides.

The limit as h → 0 is written as

0limh

2 2 2

0

2

0

0

0

2( ) lim

2( ) lim

(2 )( ) lim

( ) lim(2 )

( ) 2

h

h

h

h

x xh h xf x

h

xh hf x

hh x h

f xh

f x x h

f x x

h gets so small its effectively zero.

0

( ) ( )( ) lim

h

f x h f xf x

h

2 2

0

( )( ) lim

h

x h xf x

h

What does the answer mean?For each and every point on the

curve f(x)=x2 , the gradient of the tangent to the curve is given by the formula

f ′(x)=2x

Value of x Gradient of the tangent

-3 -6

-2 -4

-1 -2

0 0

1 2

2 4

3 6

4 8

A table of results might make this clearer

y

x

2

2

4

4

– 2

– 2

– 4

– 4

2

2

4

4

6

6

8

8

10

10

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

m = -4

y

x

2

2

4

4

– 2

– 2

– 4

– 4

2

2

4

4

6

6

8

8

10

10

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

m = -2

y

x

2

2

4

4

– 2

– 2

– 4

– 4

2

2

4

4

6

6

8

8

10

10

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

m = 0

y

x

2

2

4

4

– 2

– 2

– 4

– 4

2

2

4

4

6

6

8

8

10

10

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

m = 2

y

x

2

2

4

4

– 2

– 2

– 4

– 4

2

2

4

4

6

6

8

8

10

10

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

m =

4

Is there an easier way to do this?

y

x

2

2

4

4

– 2

– 2

– 4

– 4

2

2

4

4

6

6

8

8

10

10

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

m = -6

Rules for differentiationThere are four rules for differentiating –

remember these and you can differentiate anything …

Rule Examples

f(x) = xn f ′(x) = nxn-1 f(x) = x6 f ′ (x) =

f(x) = cxn f ′ (x) = cnxn-1 f(x)= 4x2 f ′ (x) =

f(x) = c f ′ (x) = 0 f(x) = 65 f ′ (x) =

f(x) = g(x) + h(x) f ′ (x) = g′ (x) + h′ (x)

f(x)= x6 + 4x2 + 65 f ′ (x) =

6 x 6 - 1 = 6 x 5

4 x 2 x 2-1

= 8 x 1 or 8 x

0

6 x 5 + 8x

Derivatives of ( ) nf x ax3.

The Derivative

The process of deriving '( )f x from f(x) is called differentiation.

'( )f x represents two things:

• The rate of change of the function

• The gradient of the tangent to the function

Copy the following:

multiply by the power

reduce the power by 1

Basic Rule

If f(x) = axn then f '(x) = naxn-1

Power must be rational

Example 1

Find the derivative of:

(a)9x (b)

94x

(c)5 3x 7

2

x(d)

(9 1)9x 89x

( 9 1)9 4x 1036x

Think “Flower Power”:

Power on top

Root at bottom

35x

3 55 5( )3

5 x 253

5 x

722x

7 22 2( )7

2 2x 927x

Must be if form axn

Heinemann , p.95, EX 6F, Q1-10

' ( 5 1)( ) 5 1f x x

Example 2

Find the gradient of the tangent to the curve 5

1( )f x

x

Solution:

5( )f x x

'6

5( )f x

x

'6

5(2)

(2)f

at x = 2

5

1( )f x

x

1. Prepare for differentiation (ie must be if form axn)

2. “Multiply by power then reduce power by 1”

' 6( ) 5f x x

3. Tidy up

' 5(2)

64f

4. Substitute in given value for x

Example 3

The volume in a container can be calculated using

Solution:

3 2( )V t tCalculate the rate of change of the volume after 8 seconds.

1. Prepare for differentiation (ie must be if form axn)

3 2( )V t t23( )V t t

2. “Multiply by power then reduce power by 1”

323 3( )' 2

3( )V t t 13' 2

3( )V t t 3. Tidy up '

3

2 1( )

3V t

t

4. Substitute in given value for t'

3

2 1(8)

3 8V

Example 3

The volume in a container can be calculated using

Solution:

3 2( )V t tCalculate the rate of change of the volume after 8 seconds.

1. Prepare for differentiation (ie must be if form axn)

2. “Multiply by power then reduce power by 1”

3. Tidy up

4. Substitute in given value for t

'

3

2 1(8)

3 8V

' 2 1(8)

3 2V

' 2 1(8)

6 3V

Heinemann , p.92, EX 6E

Derivatives of complex expressions

( ) ( ) ( )f x g x h x

4.

( ) ( ) ( )f x g x h x

( )( )

( )

g xf x

h x

The Derivative of Multiple Terms in x

So far we have differentiated functions with only one term in x.

' ' '( ) ( ) ( )f x g x h x

Our basic rule still applies when we have multiple terms in x.

We simply have to ensure that each individual term has been preparedfor differentiation. Then differentiate each term separately.

This means that if ( ) ( ) ( )f x g x h x

then

Example 4

Find the derivative of 4 212

3( ) 3 9

2f x x x

x

Solution:

' 3 232( ) 2 6f x x x x

1. Prepare for differentiation (ie must be if form axn)

2. “Multiply by power then reduce power by 1” for each term ' 3

2

3( ) 2 6

2f x x x

x

3. Tidy up

4 2 1312 2( ) 3 9f x x x x

' (4 1) (2 1) ( 1 1)312 2( ) 4 2 3 1 0f x x x x

Heinemann , p.94, EX 6F, Q19-27

Example 5

Solution:

1. Prepare for differentiation (ie expand brackets and simplify)

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

2( ) 6 12 9 18f x x x x

2. “Multiply by power then reduce power by 1”

3. Tidy up

Find the derivative of f(x) = (2x – 3) (3x + 6)

2( ) 6 3 18f x x x

'( ) 12 3f x x

Heinemann , p.95, EX 6G, Q 1-6

Example 6

Solution:

2. Prepare for differentiation (ie use laws of indices)

3. “Multiply by power then reduce power by 1”

4. Tidy up

1. Re-write expression with each term in the numerator over the denominator

Find the derivative of 3 2

2

4 3( )

x x xf x

x

3 2

2 2 2 2

4 3( )

x x xf x

x x x x

1 2( ) 4 3f x x x x

' 2 3( ) 1 6f x x x

'2 3

1 6( ) 1f x

x x

NAB

Heinemann , p.95, EX 6G, Q 17-24

Leibniz Notation

2dy ax bx cdx

5.NewtonLeibniz

The last years of his life - from 1709 to 1716 - were embittered by the long controversy with John Keill, Newton, and others, as to whether he had discovered the differential calculus independently of Newton's previous investigations, or whether he had derived the fundamental idea from Newton, and merely invented another notation for it. The controversy occupies a place in the scientific history of the early years of the eighteenth century quite disproportionate to its true importance, but it materially affected the history of mathematics in western Europe .

Gottfried Wilhelm Leibnitz (1646 - 1716)

On the accession in 1714 of his master, George I.,to the throne of England, Leibnitz was thrown asideas a useless tool; he was forbidden to come to England; and the last two years of his life were spent in neglect and dishonour. He was overfond of money and personal distinctions; was unscrupulous, as perhapsmight be expected of a professional diplomatist of thattime; but all who once came under the charm of hispersonal presence remained sincerely attached to him.

He also held eminent positions in diplomacy, philosophy and literature.

From `A Short Account of the History of Mathematics' (4th edition, 1908) by W. W. Rouse Ball.

Place in History

1660 Charles ii restores English monarchy

1665/6 Plague & Great fire of London

1689 William of Orange takes English Crown

1692 Massacre in Glencoe

1694 Bank of England Formed1695 Bank of Scotland Formed

1707 Act of Union

1712 First steam engine

1714 George I begins Hanovarian dynasty leadingto Robert Walpole becoming first Prime Minister.

1666 Newton’s correspondence refers to “fluxions”and “the sum of infinitesimals”

1675 Leibniz makes the calculus public (published 1684)

1693 Newton publishes work on the calculus

Leibniz NotationLiebniz Notation is an alternative way of expressing derivatives to f'(x) , g'(x) , etc.

If y is expressed in terms of x then the derivative is written as dy/dx .

eg y = 3x2 - 7x so dy/dx = 6x - 7 .

Regardless of the notation, the meaning of the result is the same:

• the rate of change of the function

• the gradient of the tangent at a given point

“dee y by dee x”

If function is given as y=, or x=, t= etc. we use Leibniz notation.

Example 1

Solution:

1. Prepare for differentiation2 39 15Q R R

(2 1) ( 3 1)2 9 ( 3) 15dQ

R RdR

2. “Multiply by power then reduce power by 1”

3. Tidy up 418 45dQ

R RdR

4

4518

dQR

dR R

If Q = 9R2 - 15 R3

find dQ/dR !

A curve has equation y = 5x3 - 4x2 + 7 . Find the gradient of the tangent at x = -2

Example 2

Solution:

1. Prepare for differentiation (NB. y =, so Leibniz notation)

2. “Multiply by power then reduce power by 1”

3. Substitute in given x-value

3 25 4 7y x x

215 8dy

x xdx

215( 2) 8( 2)( 2)

dy

dx

76( 2)

dy

dx

60 16( 2)

dy

dx

Gradient at x = -2 is 76

4. Communicate that you have answered question.

Extra bit for “FIZZY SYSTS” or even Physicists.

Newton’s 2ndLaw of Motion

s = ut + 1/2at2 where s = distance & t = time.

Finding ds/dt gives us a “diff in dist” “diff in time”

ie speed or velocity

so ds/dt = u + at

but ds/dt = v so we get v = u + at

and this is Newton’s 1st Law of Motion

Heinemann , p.100, EX 6I, Q 2-5

The Equation of the tangent

'( )dy

f xdx

6.

1 1( )y y m x x

The Tangent to the Graph

So far we have differentiated functions, stating that the derivedfunction provides us with the gradient of the tangent to the graph.

tangent

y = f(x)

A (a,b)

y = mx + c

We have seen that whilst the derived function is the same throughoutthe curve the actual gradient is dependant on the x-coordinate.

The Tangent to the Graph

Now we want to find the equation of the tangent.

tangent

y = f(x)

A (a,b)

y = mx + c

The tangent is a straight line. We know from previous work on thestraight line that . How can we find m?1 1( )y y m x x

The Tangent to the Graph

At any point gradient of curve = gradient of tangent

tangent

y = f(x)

A (a,b)

y = mx + c

As the gradient of a straight line is the same throughout the line if we can find the gradient of the curve at A this is the gradient we can use for the equation of the tangent. So

'( )m f a

Finding the Equation of the Tangent

To find the equation of the tangent we:

1. Must have both coordinates of the point on the curve

2. Must have the gradient at that point.

3. Substitute these details into: 1 1( )y y m x x

Copy the following:

Example 1

Find the equation of the tangent to the curve 45 1y x

Solution:

320dy

xdx

1. If not given, find y-coordinate

2. Find dy / dx

320(1)(1)

dy

dx3. Find gradient at x-coordinate

6 20( 1)y x

at x = 1

45 1y x

45(1) 1y =6

= 20

4. Substitute values into 1 1( )y y m x x

6 20 20y x

20 14y x

Equation of tangent at (1,6):

NAB

Example 2

Solution:

1. Note we are being asked to prove that dy/dx = 36 has only one solution.

10 6dy

xdx

36 10 6x 2. Find dy/dx

3. Make an equation with given gradient and solve

Show that there is only one tangent to the curve

10 30x 3x

25 6y x x with gradient 36

4. Make statement As dy/dx = 36 has only one solution there is only onetangent with that gradient.

2 6

3

x

x

Example 3

Solution:

2. Make dy / dx equal to given gradient and solve.

3. Substitute into original equation to find y-coordinate. (DO NOT USE dy / dx.)

4. Make statement

1. Find dy / dx

Find the point of contact of the curve 2 3 8y x x

2 3dy

xdx

2 3 9x

2(3) 3(3) 8y

at which thegradient is 9.

2 3 8y x x

y-coord when x = 3:

9 9 8 10y

Point of contact (3 , 10)

Heinemann , p.101, EX 6J,Q1, 3, 5, 6 & 7

Increasing / Decreasing Functions

'( ) 0f x

7.

'( ) 0f x

Is the function increasing or decreasing?

Our next task is to establish whether or not a function is increasingor decreasing.

tangent

y = f(x)

A (a,b)

y = mx + c

What can we say about the gradient of the tangent as the y- coordinatesincrease i.e. the value of the function is increasing?

Whilst f(x) is increasinggradient is POSITIVE.

Is the function increasing or decreasing?

y = f(x)

What can we say about the gradient of the tangent as the y- coordinatesdecrease i.e. the value of the function is decreasing?

Whilst f(x) is decreasinggradient is NEGATIVE.

Is the function increasing or decreasing?

What happens when the graph is changing direction?

y = f(x)

The tangent is horizontal and so the gradient is ZERO.

These are called stationary points.

Whilst f(x) is not changinggradient is ZERO

Increasing and Decreasing Functions

y = f(x)

Copy the following:

'( ) 0f x '( ) 0f x

'( ) 0f x '( ) 0f x Function increasing

Function decreasing'( ) 0f x '( ) 0f x Stationary Point

Example 1

State whether the function 3 24 3 3y x x

Solution:212 6

dyx x

dx 1. Find dy / dx

212 ( 2) 6 ( 2)( 2)

dy

dx

2. Find gradient at x-coordinate

is increasing or

= 36

3. Make statement

As dy/dx is positive, functionis increasing at x = -2

decreasing at x = -23 24 3 3y x x

48 12( 2)

dy

dx

Example 2

Solution:

1. Find dy/dx 212 12dy

x xdx

212 12 0x x 2. Set dy/dx = 0 and solve to find x-coords of Stationary Points

3. Make a table to show what gradient is before and after SP’s

Find the intervals in which the function

12 ( 1) 0x x

3 24 6 2y x x is increasing and decreasing.

3 24 6 2y x x

Factorise

1 0x

1x

12 0x

0x

or

or

X -1 0

dy

dx0 0 ++ -

Example 2

Solution:

4. Make statement

Find the interval in which the function 3 24 6 2y x x is increasing and decreasing.

X -1 0

dy

dx0 0 ++ -

f(x) increasing:

0x

1x

f(x) decreasing: 1 0x

f(x) increasing:

The Proof3 24 6 2y x x f(x) increasing:

0x

1x

f(x) decreasing: 1 0x

f(x) increasing:

Example 3

Solution:

2. Find dy / dx

3. Factorise (all terms involving x must have even powers).

4. Make statement

1. Prepare for differentiation

Show that the function3

23 9 43

xy x x

2 6 9dy

x xdx

2( 3)dy

xdx

is never decreasing.

3 213 3 9 4y x x x

Since (x - 3)2 will always produce a positive gradientfunction is never decreasing.

323 9 4

3

xy x x

The Proof3

23 9 43

xy x x

Heinemann , p.104, EX 6L, Q1 to 6

Stationary Points

'( ) 0f x

8.

What if the function is neither increasing nor decreasing?

Recall from the last lesson that when the gradient is changing there will be points where

y = f(x)

the tangent is horizontal and so the gradient=ZERO.

These are called stationary points.

Whilst f(x) is not changinggradient is ZERO

At these points '( ) 0f x

The Nature of a Stationary Point Copy the following:

The nature of a stationary point is determined by the gradient eitherside of it. There are four classifications:

Maximum Turning Point Minimum Turning Point

The Nature of a Stationary Point Copy the following:

The nature of a stationary point is determined by the gradient eitherside of it. There are four classifications:

Rising Point of Inflection Falling Point of Inflection

Example 1

Determine the coordinates and nature of the stationary points on the curve 35 ( 4)y x x

Solution:

3 220 60dy

x xdx

2. Find dy / dx

3 20 20 60x x 3. Set dy/dx equal to zero, factorise if possible, and solve for x

4 35 20y x x

20 20 ( 3)x x

35 ( 4)y x x

1. Prepare for differentiation

220 0x or ( 3) 0x

x = 0 or x = 3

At SP’s dy/dx = 0

In exams must make this statement

NAB

Example 1

Determine the coordinates and nature of the stationary points on the curve 35 ( 4)y x x

Solution:

5. State coords of Stationary Points

4. Find y-coordinates by subbing these values into original equation (NOT dy/dx)

x = 0 or x = 3

For x = 0:

35(0) ((0) 4)y

0y

For x = 3:

35(3) ((3) 4)y

5 27 1y

135y

SP’s are (0,0) and (3,-135)

Example 1

Determine the coordinates and nature of the stationary points on the curve 35 ( 4)y x x

Solution:

7. Make statement

6. Draw a NATURE TABLE for each SP to determine its nature.

SP’s are (0,0) and (3,-135)

X 0

-ve 0 -ve

Slope

3 220 60dy

x xdx

dy

dx

0 0

(0,0)

(0,0) is a falling point of inflexion.

3 220( 1) 60( 1) 80( 1)

dy

d

3 220(1) 60(1) 40(1)

dy

d

Example 1

Determine the coordinates and nature of the stationary points on the curve 35 ( 4)y x x

Solution:

7. Make statement

6. Draw a NATURE TABLE for each SP to determine its nature.

SP’s are (0,0) and (3,-135)

X 3

-ve 0 +ve

Slope

3 220 60dy

x xdx

dy

dx

3 3

(3,-135)

(3,-135) is a minimumturning point.

3 220(1) 60(1) 40(1)

dy

d

3 220(4) 60(4) 320(4)

dy

d

The Proof35 ( 4)y x x

Heinemann , p.106, EX 6M

Curve Sketching9.

What do we need to know to sketch the curve?

If asked to sketch this curve what information do you think you wouldneed?

y = f(x)Stationary Points

y-intercept

x-intercept

Behaviour as x gets small

Behaviour as x gets big

Curve Sketching Copy the following:

If asked to sketch a curve we need to establish the following:

1. y-intercept (x = 0)

2. x-intercept(s) (y = 0)

3. Stationary Points (dy/dx = 0 )

4. Behaviour as x gets very big and very small x x

Example 1

Sketch the graph of 3 48 3y x x

Solution:

3 40 8 3x x

3 48(0) 3(0)y

30 (8 3 )x x

3 48 3y x x

3 0x or (8 3 ) 0x

1. Find y-intercept (x = 0)= 0

y-intercept is (0,0)

2. Find x-intercept (y = 0)

3 8x 8

3x

x-intercepts (0,0) & ( ,0)232

x = 0

Example 1

Sketch the graph of 3 48 3y x x

Solution: 2 324 12dy

x xdx

20 12 (2 )x x

3 48 3y x x

212 0x or (2 ) 0x

x = 0 x =2

At SP’s dy/dx = 03. Find Stationary Points and their nature (dy/dx = 0 )

2 30 24 12x x

3 48(0) 3(0)y

y = 0

3 48(2) 3(2)y y = 16

X 0 2

0 0

Slope

Example 1

Sketch the graph of 3 48 3y x x

Solution:

dy

dx

2 324 12dy

x xdx

3. Find Stationary Points and their nature (dy/dx = 0 )

SP’s are (0,0) and (2,16)

0 2

+ve+ve -ve

(0,0) is a rising point of inflexion

(2,16) is a maximum tp

2 324( 1) 12( 1) 36( 1)

dy

d

2 324(1) 12(1) 12

(1)

dy

d

2 324(3) 12(3) 108(3)

dy

d

Example 1

Sketch the graph of 3 48 3y x x

Solution: 3 48 3y x x

4. Behaviour as x gets very big and very small x

x As x -3x4 will dominate

and so large x’s will be -ve

As x -3x4 dominates

and so small x’s will be -ve

Even Power - +veOdd Power – get what you started with

5. Plot the SP’s, the intercepts and draw curve.

Biggest power

dominates

Solution: y-intercept is (0,0) x-intercepts (0,0) & ( ,0)232

(0,0) is a rising point of inflexion (2,16) is a maximum tp

(2,16)

(0,0)

232

Solution: y-intercept is (0,0) x-intercepts (0,0) & ( ,0)232

(0,0) is a rising point of inflexion (2,16) is a maximum tp

(2,16)

(0,0)

232

3 48 3y x x

Heinemann , p.107, EX 6NQ1, 2, 5, 7 & 8

Max /Min on Closed interval

10.

What is a closed interval?

y = f(x)

Up to now we have looked at graphs without any restrictions i.e wehave considered the function and its graph for all possible values of x.

Sometimes we may wish to just look at a certain part of the function or graph i.e restrict the interval.

Closed interval

Where are the maximum and minimum values in a closed interval?

Lets consider f(x) = 8 + 2x – x2 on the interval 1 2x

-1 5

-0.9 5.39

-0.8 5.76

-0.7 6.11

-0.6 6.44

-0.5 6.75

-0.4 7.04

-0.3 7.31

-0.2 7.56

-0.1 7.79

0 8

0.1 8.19

0.2 8.36

0.3 8.51

0.4 8.64

0.5 8.75

0.6 8.84

0.7 8.91

0.8 8.96

0.9 8.99

1.1 8.99

1.2 8.96

1.3 8.91

1.4 8.84

1.5 8.75

1.6 8.64

1.7 8.51

1.8 8.36

1.9 8.19

2 8

x f(x)f(x)

f(x)x

xSmallest 1 9Largest

& SP

The proof

Maximum & Minimum Values on a Closed Interval

In a closed interval the maximum and minimum values of a function are either at a stationary point or at

an end point of the interval.

y

x

Local maximum

Local minimum

A closed interval is often written as { −2 < x < 3} or [−2 , 3]

This interval would refer to the values of the function from −2 to 3

Copy the following:

Example 1

Find the maximum and minimum values of 2 34 3y x x

Solution:

20 6 3x x

2 34 3( 2) ( 2)y

0 3 (2 )x x

26 3dy

x xdx

3 0x or (2 ) 0x

1. Find Stationary Points

2x

SP’s are (0,4) & (-2,0)

x = 0

within the interval 3 1x 2 34 3y x x

At SP’s dy/dx = 0

y = 4y = 0

Example 1

Find the maximum and minimum values of 2 34 3y x x

Solution:

2. Find value of f(x) at each end point of given interval

within the interval 3 1x 2 34 3y x x

At x = -3

2 34 3( 3) ( 3)y 4 3(9) ( 27)y

4 27 27y

y = 4 (-3,4)

Example 1

Find the maximum and minimum values of 2 34 3y x x

Solution:

2. Find value of f(x) at each end point of given interval

within the interval 3 1x 2 34 3y x x

At x = 1

2 34 3(1) (1)y 4 3(1) (1)y

4 3 1y

y = 0

3. Compare Y-COORDINATES to find smallest and biggest

(0,4) & (-2,0)

(1,0)

(-3,4) (1,0)

4. Make statement Maximum = 4, Minimum = 0

The Proof

Heinemann , p.109, EX 6OQ2 (a), (b), (e) & (f)

Graph of the Derived Function

11.

What will the graph of a derivative look like?

y = x3 – 6x2 + 9x

Imagine you were asked to sketch a graph of '( )f x

What would the key points be?

Gradient +veto SP 1

Gradient –veto SP 2 Gradient

+ve

Is the gradient constant?

Lets consider dy/dx = 3x2 – 12x + 9

x dy/dxdy/dx

dy/dxx

x

y = x3 – 6x2 + 9x

-1 24

-0.75 19.7

-0.5 15.8

-0.25 12.2

0 9

0.25 6.19

0.5 3.75

0.75 1.69

1 0

1.25 -1.31

1.5 -2.25

1.75 -2.81

2 -3

2.25 -2.81

2.5 -2.25

2.75 -1.31

3 0

3.25 1.688

3.5 3.75

3.75 6.188

4 9

4.25 12.19

4.5 15.75

4.75 19.69

5 24

Moving to zero

SP at (1,0)

Min = -3

Decreasing to (2,-3) then increasing

SP at (3,0)

The original and the derived

y = f(x)'( )y f x

Example 1

Sketch the derived function for f(x)y = f(x)

SP occurs at x = 5

x

f '(x)

5

- 0 + New y-values

Example 1

SP occurs at x = 5

x

f '(x)

5

- 0 + New y-values

Negative butapproaching zero

Example 1

SP occurs at x = 5

x

f '(x)

5

- 0 + New y-values

Negative butapproaching zero

Positives

'( )y f x

Example 2

Sketch the derived function for f(x)

(2,-4)

SPs occur at x = 0 & x = 2

x

f'(x)

0 2

+ 0 - 0 + New y-values

y = f(x)

Example 2

Sketch the derived function for f(x)

(2,-4)

SPs occur at x = 0 & x = 2

x

f'(x)

0 2

+ 0 - 0 + New y-values

Example 2

Sketch the derived function for f(x)

(2,-4)

SPs occur at x = 0 & x = 2

x

f'(x)

0 2

+ 0 - 0 + New y-values

Positive butapproaching zero

Example 2

Sketch the derived function for f(x)

(2,-4)

SPs occur at x = 0 & x = 2

x

f'(x)

0 2

+ 0 - 0 + New y-values

Goes into –ve’sbut returns to zero by 2

Example 2

Sketch the derived function for f(x)

SPs occur at x = 0 & x = 2

x

f'(x)

0 2

+ 0 - 0 + New y-values

Into +ve’s

Heinemann , p.111, EX 6P

Graph of the Derived Function

11.

What will the graph of a derivative look like?

y = x3 – 6x2 + 9x

Imagine you were asked to sketch a graph of '( )f x

What would the key points be?

Gradient +veto SP 1

Gradient –veto SP 2 Gradient

+ve

Is the gradient constant?

Lets consider dy/dx = 3x2 – 12x + 9

x dy/dxdy/dx

dy/dxx

x

y = x3 – 6x2 + 9x

-1 24

-0.75 19.7

-0.5 15.8

-0.25 12.2

0 9

0.25 6.19

0.5 3.75

0.75 1.69

1 0

1.25 -1.31

1.5 -2.25

1.75 -2.81

2 -3

2.25 -2.81

2.5 -2.25

2.75 -1.31

3 0

3.25 1.688

3.5 3.75

3.75 6.188

4 9

4.25 12.19

4.5 15.75

4.75 19.69

5 24

Moving to zero

SP at (1,0)

Min = -3

Decreasing to (2,-3) then increasing

SP at (3,0)

The original and the derived

y = f(x)'( )y f x

Example 1

Sketch the derived function for f(x)y = f(x)

SP occurs at x = 5

x

f '(x)

5

- 0 + New y-values

Example 1

SP occurs at x = 5

x

f '(x)

5

- 0 + New y-values

Negative butapproaching zero

Example 1

SP occurs at x = 5

x

f '(x)

5

- 0 + New y-values

Negative butapproaching zero

Positives

'( )y f x

Example 2

Sketch the derived function for f(x)

(2,-4)

SPs occur at x = 0 & x = 2

x

f'(x)

0 2

+ 0 - 0 + New y-values

y = f(x)

Example 2

Sketch the derived function for f(x)

(2,-4)

SPs occur at x = 0 & x = 2

x

f'(x)

0 2

+ 0 - 0 + New y-values

Example 2

Sketch the derived function for f(x)

(2,-4)

SPs occur at x = 0 & x = 2

x

f'(x)

0 2

+ 0 - 0 + New y-values

Positive butapproaching zero

Example 2

Sketch the derived function for f(x)

(2,-4)

SPs occur at x = 0 & x = 2

x

f'(x)

0 2

+ 0 - 0 + New y-values

Goes into –ve’sbut returns to zero by 2

Example 2

Sketch the derived function for f(x)

SPs occur at x = 0 & x = 2

x

f'(x)

0 2

+ 0 - 0 + New y-values

Into +ve’s

Heinemann , p.111, EX 6P

Optimisation : Maxima & Minima

12.

Problem Solving

Differentiation can be used to solve problems which require maximum or minimum values.

Problems typically cover topics such as areas, volumes and rates of change. They often involve having to establish a suitable formula in one variable and then differentiating to find a maximum or minimum value. This is known as Optimisation.

It is important to check the validity of any solutions as often an answer is either nonexistent (e.g. a negative length or time) or outside an acceptable interval.

OptimisationOptimisation is the term we use to describe a simple process of finding a maximum or minimum value for a given situation. Normally we would have to graph functions and then use our graph to establish a maximum or minimum.

As we have learned previously, differentiation allows us to quickly find the required value and is the expected manner of solving problems like this in Higher mathematics.

Drawing graphs would not be an acceptable solution.

OptimisationProblems posed will often involve more than one variable. The process of differentiation requires that we rewrite or re-arrange formulae so that there is only one variable – typically x.

Questions are therefore multi-part where the first part would involve establishing a formula in x from a given situation, part 2 would involve the differentiation and validation of an acceptable answer, with part 3 the solution to the problem set.

You would do well to note that although the first part is important, you can normally expect to complete the rest of the question even when you cannot justify a formula in part 1.

Optimization : Maxima and Minima

•Differentiation is most commonly used to solve problems by providing a “best fit” solution.

•Maximum and minimum values can be obtained from the Stationary Points and their nature.

•In exams you may be asked to “prove” a particular formula is valid. Even if you cannot prove this USE THIS FORMULA TO ANSWER THE REST OF THE QUESTION.

Copy the following:

Example 1 (Formula Given)

The “Smelly Place” Garden Centre has a model train which is usedto show visitors around the many displays of plants and flowers.

It has been established that the cost per month, C, of running the train is given by:

40050 25C V

V

where V is the speed in miles per hour.

Calculate the speed which makes the cost per month a minimum andhence calculate this cost.

Example 1 40050 25C V

V

Solution:

1. Find derivative, rememberingto prepare for differentiation.

150 25 400C V V

225 400dC

VdV

2

40025

dC

dV V

Example 1 40050 25C V

V

Solution:

2. Make statement thenset derivative equal to zero.

2

400 25

1V

2

4000 25

V

2

40025

dC

dV V

At SP’s dC/dV = 0

Cross Multiply

2400 25V2 16V

4V

Reject –ve speed

Example 1 40050 25C V

V

Solution:

3. Justify nature of each SPusing a nature table.

2

40025

dC

dV V 4V

X

Slope

dC

dV

44 4

-ve 0 +ve

A minimum tp occurs and so costis minimised at V = 4.

4. Make statement

2

40025 375

(1) (1)

dC

d

2

40025 17

(5) (5)

dC

d

Example 1 40050 25C V

V

Solution:

4. Sub x-value found intooriginal expression to find corresponding minimum value.

A minimum tp occurs and so costis minimised at V = 4.

40050 25C V

V

40050 25(4)

4C

For V = 4:

50 100 100C C = £250

5. Answer QuestionMinimum cost is £250 per monthand occurs when speed = 4mph

(DO NOT USE dy/dx)

Heinemann , p.115, EX 6RQ3

Example 2 (Perimeter and area)

A desk is designed which is rectangular in shape and which is requiredin the design brief to have a perimeter of 420 cm.

If x is the length of the base of the desk:

(a) Find an expression for the area of the desk, in terms of x.

(b) Find the dimensions which will give the maximum area.

(c) Calculate this maximum area.

Example 2

Solution to (a):

1. Start with what you know

2. In order to differentiate we need expression only involveX’s. Use other information tofind an expression for Y. 420 2 2x y

Area = length x breadth

Lets call breadth y, so: Area = x x y

Perimeter = 2x + 2y

3. Change subject to y.2 420 2y x

210y x

Divide by 2

x

y

Example 2

Solution to (a):

4. Now substitute this expressionfor y into our original expressionfor AREA.

( ) 210 2A x x

2( ) 210Area A x x x

At SP’s A’(x) = 0

210 2 0x

2 210x

105x

Area = x x y 210y x

Area = x x (210 – x)

5. To find dimensions givingmaximum or minimum first findderivative.

6. Make statement and set derivative equal to zero.

Solution to (b):

Example 2

Solution to (a):

7. Now use your expression for y to find other dimension.

105x

Area = x x y 210y x

210 105 105y

Example 2

Solution:

8. Must justify maximumvalues using nature table.

X

Slope

( )A x

105

+ve 0 -ve

The area is maximised whenlength is 105cm and breadth is 105 cm.

9. Make statement

( ) 210 2A x x

X = 105 and Y = 105

(100) 210 2(100) 10A

(110) 210 2(110) 10A

Example 2

Solution to (c):

10. To find actual maximum areasub maximum x-value just found into expression for area.

max 22,050 11,025Area (DO NOT USE dy/dx)

The area is maximised whenlength is 105cm and breadth is 105 cm.

2( ) 210Area A x x x

2max 210(105) (105)Area

2max 11,025Area cm

PROOF

Heinemann , p.112, EX 6QQ1, 2, 5

Example 3 (Surface Area and Volume)

A cuboid of volume 51.2 cm3 is made with a length 4 times itsbreadth.

If x cm is the breadth of the base of the cuboid:

(a) Find an expression for the surface area of the cuboid, in terms of x.

(b) Find the dimensions which will give the minimum surface areaand calculate this area.

xcm

Example 3

Solution to (a):

1. Start with what you know

2. In order to differentiate expression must only involveX’s. Use other information tofind an expression for h.

251.2 4x h

Surface Area = sum of area of faces

Lets call height h, so: = 2(4x x h)

Volume = 4x x x x h

3. Change subject to h. 2 2

51.2 12.8

4h

x x

x4x

h

+ 2(x x h) + 2(4x x x)

= 10xh + 8x2

4. Now substitute this expressionfor y into our original expressionfor SURFACE AREA.

Example 2

Solution to (a):

22

12.8( ) 10 8A x x x

x

Surface Area = sum of area of faces

2

12.8h

x

22

128( ) 8

xA x x

x

2 128( ) 8A x x

x

= 10xh + 8x2

Example 3

Solution to (b):

5. To find dimensions giving maximum or minimum first find derivative.

2 128( ) 8A x x

x

2

128( ) 16A x x

x

2 1( ) 8 128A x x x

6. Make a statement and set derivative equal to zero.

At SP’s A’(x) = 0

2

12816 0x

x

316 128 0x

Multiplyby x2

316 128x 3 8x

3 8 2x

Example 3

Solution to (b):

7. To find corresponding height substitute x value just found into expression for h found in step 2. So dimensions minimising area:

2

12.8 12.83.2

(2) 4h

2

12.8h

x

Breadth = 2 cm

Length = 8 cm

Height = 3.2 cm

Example 3

Solution:

8. Must justify minimumvalues using nature table.

X

Slope

( )A x

2

-ve 0 +ve

The surface area is minimised whenthe breadth is 2 cm

9. Make statement

(1) 16 128 112A

128(4) 16(4) 56

16A

2

128( ) 16A x x

x

3 8 2x

Example 3

Solution to (c):

10. To find actual maximum areasub maximum x-value just found into expression for area.

min 32 64Area (DO NOT USE dy/dx)

2min

1288(2)

2Area

2min 96Area cm

2 128( ) 8A x x

x

The surface area is minimised whenthe breadth is 2 cm

PROOF

Heinemann , p.112, EX 6RQ1, 2, 5

Example 4 ( ADDITIONAL - NOT ESSENTIAL)

A channel for carrying cables is being dug out at the side of the road.

A flat section of plastic is bent into the shape of a gutter and placedInto the channel to protect the cables.

40

cm

100 cm

x

x

The dotted line represents the fold in the plastic, x cm from either end.

(a) Show that the volume of each section of guttering is ( ) 200 (20 )V x x x

(b) Calculate the value of x which gives the maximum volume of gutter and find this volume.

Example 2( ) 200 (20 )V x x x

Solution to part (a):

Require volume so:

V l b h Length = 100 cm

Breadth = (40 – 2x) cm

Height = x cm100 (40 2 )V x x

100 (40 2 )V x x

100 2(20 )V x x

200 (20 )V x x

Factorise bracket

(a) Show that the volume of each section of guttering is

“folded bit” =

as required.

Example 2 ( ) 200 (20 )V x x x

Solution to part (b):

'( ) 4000 400V x x

Maximum and minimum will occur at SP’s

(b) Calculate the value of x which gives the maximum volume of gutter.

( ) 200 (20 )V x x x

2( ) 4000 200V x x x Must prepare for differentiation

At SP’s '( ) 0V x

0 4000 400x 0 400(10 )x 0 (10 )x x = 10 Now prove this is a maximum tp

Must make this statement

Example 2 ( ) 200 (20 )V x x x

Solution to part (b):

(b) Calculate the value of x which gives the maximum volume of gutter.

Now prove this is a maximum tpx = 10

X 10

0

Slope

dV

dx

10 10

+ve -ve

'( ) 4000 400V x x

So when x = 10 we have a maximum

Example 2 ( ) 200 (20 )V x x x

Solution to part (b):

(b) Calculate the value of x which gives the maximum volume of gutter.

Now calculate maximum volume( ) 200 (20 )V x x x

So when x = 10 we have a maximum

(10) 200 10 (20 10)V

3(10) 20000V cm

So maximum volume of gutteringis 20,000 cm3.

Make statement