Instantaneous Speed Calculation

Higher Unit 1

Differentiation

Introduction to
1.
differentiation

Introduction to Differential Calculus

Do not go from speed x to zero
instantaneously.

Up to this moment, your experience of speed and the relationship
with distance and time has been restricted to questions such as those
represented 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.
2. The graph doesn’t realistically depict changes in speed

Therefore we will look at methods for finding the instantaneous
Speed 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.
“If I have seen further it is by
standing on the shoulders of giants.”
50
40 Sir Isaac Newton(1642-1727)
30
20
10

0 1 2 3

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:

Distance
Speed 
Time
Calculate the average speed between 0 and 3 seconds.

45  0 So far so
Speed  = 15m/s good for our
30 dastardly
duo.

So far we have worked out the average speed
between 0 and 3 seconds as shown below:
f=at2
Now repeat for 1 to 3
seconds.
50
And 2 to 3 seconds
40
45m
30
20
10

0 1 2 3

3s

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 25m/s
15m/s 20m/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. 33.8m 36.45m 42.04m 45m
39.2m

Time 2.6-3.0 2.7-3.0 2.8-3.0 2.9-3.0
Interval
Average 28.5m/s 29m/s 29.6m/s
28m/s
Speed.

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:

50
40
30
20
10

0 1 2 3

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:

The tangent to the curve
50
at t = 3 .
40
30
20
10

0 1 2 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 !!

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 …..

Rules of indices
Rule Examples
1
a0 = 1 12.3140 =
1 1
1  3
a= x= 5
-m -5
nn
3
x
am
2
m 6
=
x
x
3 2 3
am  a
n n
y 55
y6
3 1
= 6a a
2a × 3a
3/2 1/2 2 2
am × an = am+n 2
6a
=
x2 x -3= x 23
a a =a
m n m-n
x5
=
q2 ﾴ q2 ﾴ q2
(q ) =
23
(am)n = amn 6
=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 toofcurves or slope of a
The derivative function is a measure the gradient
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
B(x2,y2)
The gradient of AB = mAB
y2 – y1
Change in y

Change in x
A(x1,y1)
x2 – x1
y2  y1
C 
x 2  x1
y2  y1
m , Gradient Formula
x 2  x1

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 B
is 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.

Gradient of a function
y = f(x)
y

A
(( x  h), f ( x  h))
f ( x  h)
( x  h) y2  y1
f ( x  h) M AB 
( x  h) x2  x1
(( x), f ( x))
f ( x  h)
( x) B f ( x) ( x  h)
x ( x  ( x  h)
f
x  h f ( x) hf)( x) f ( x)
( x) x f ( x)
f ( x) f ( x) f ( x  h) f ( x )
( x)
( x  h)
f ( x  h)  f ( x )
( x)
M AB  ( x  h)
( x)
 ( x)
( x)
( x)
( x) ( x)
( x) ( x)

h
What happens if we make smaller?
f ( x  h)  f ( x )

M AB
y = f(x)
( x  h)  x
y

A
(( x  h), f ( x  h))
When x  h  x
Then gradient of
(( x), f ( x)) AB is the
B gradient of the
tangent at B
x  x xhxhxhxhxhhh
h xhxhx hx xhh 
xx  hx x
xh
x

Shrinking the tangent line is the same
as letting h get very small
y = f(x)
y
The gradient
doesn't change
as the line gets
smaller and so
the gradient at
B must be the
B same as the
gradient of the
line

x

What are we saying?

We know

f ( x  h)  f ( x )

M AB
( x  h)  x
But

f ( x  h)  f ( x )
M  lim
B h 0 ( x  h)  x

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

Putting it together…. f ( x)
f ( x  h)
f ﾢ x)  lim
(
Differentiate the function f(x) = x2
h
hﾮ 0
from first principles.
( x  h) 2  x 2
f ﾢ x)  lim
(
h
hﾮ 0
The derivative is the same as the gradient
of the tangent to the curve so we can
go straight to the gradient formula we
x 2  2 xh  h 2  x 2
f ﾢ x)  lim
saw in the previous slides. (
h
hﾮ 0

2 xh  h 2
f ﾢ x)  lim
(
h
hﾮ 0

h(2 x  h)
f ﾢ x)  lim
(
h
hﾮ 0
h gets so
f ﾢ x)  lim(2 x  h) small its
(
hﾮ 0
effectively
The limit as h → 0 is
f ﾢ x)  2 x
( zero.
written as
lim
hﾮ 0

What does the answer mean?
For each and every point on the
curve f(x)=x2 , the gradient of the
Value of x Gradient of the tangent
tangent to the curve is given by
-3 -6
the formula
-2 -4
f ′(x)=2x y
y
y
-1 -2
10
10
10

0 0
8
8
8
m=

1 2
6
6
6
-6

4
2 4
=4

4
4
m

A table of results 2might make this
=

m

3 6
m 2
-4

2
=2
=
clearer -2
4 8
m=0 m x
–4 –2 2 4 x
x
–4 –2 2 4
–4 –2 2 4
–2
–2
–2

Is there an easier way to do
–4
–4
–4

this?
–6
–6
–6
–8
–8
–8
– 10
– 10
– 10

Rules for differentiation
There 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) = 6 x 6-1 = 6 x 5

f(x)= 4x2  f ′ (x) = 4 x 2 x 2-1
 f ′ (x) = cnx
f(x) = cx n n-1

= 8 x 1 or 8 x
f(x) = c  f ′ (x) = 0 f(x) = 65  f ′ (x) = 0

f(x) = g(x) + h(x)
f(x)= x6 + 4x2 + 65  f ′ (x) = 6 x 5 + 8x
 f ′ (x) = g′ (x) + h′ (x)

Derivatives of f ( x)  ax
n
3.

Copy the following:
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
Power must
• The gradient of the tangent to the function be rational
Basic Rule

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

multiply by reduce the
the power power by 1

Example 1

Find the derivative of:
9
4x
9
x (b)
(a)
Think “Flower Power”:
 9 ﾴ 4x
(9 1) ( 9 1)
 9x
10
 9x  36x
8
Power on top

2
Root at bottom
5 3 (d)
x 7
(c) x
7
3

 2x
x 5 2
Must be if form axn
(55)
3
5x 5
3 ( 7  2 )
  ﾴ 2x
7 22
2
5
2
9
x  7x
3 2
5

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

Example 2
1
Find the gradient of the tangent to the curve f ( x)  5 at x = 2
x
1
f ( x)  5
Solution:
x
f ( x)  x 5
1. Prepare for differentiation
(ie must be if form axn)
f ' ( x)  5 ﾴ 1x ( 51)
2. “Multiply by power then
f ' ( x)  5 x 6
reduce power by 1”
5
f ( x)  6
'

x
3. Tidy up
5
f (2) 
'

(2)6
4. Substitute in given value for x
5
f (2)  
'

64

Example 3

The volume in a container can be calculated using V (t ) 
3 2
t
Calculate the rate of change of the volume after 8 seconds.

V (t )  t 3 2
Solution:
2
V (t )  t 3
2 ( 33)
23
V (t )  t
'
2. “Multiply by power then 3
1
V (t )  t
' 2 3
3
21
3. Tidy up V (t )  ﾴ 3
'

3 t
4. Substitute in given value for t 21
V (8)  ﾴ 3
'

3 8

Example 3

The volume in a container can be calculated using V (t ) 
3 2
t
Calculate the rate of change of the volume after 8 seconds.
21
V (8)  ﾴ 3
'
Solution:
3 8
(ie must be if form axn) 21
V (8)  ﾴ
'

32
reduce power by 1” 21
V (8)  '

63
3. Tidy up

4. Substitute in given value for t

Derivatives of complex
4.
expressions

f ( x )  g ( x ) ﾴ h( x )
f ( x )  g ( x )  h( x )
g ( x)
f ( x) 
h( x )

The Derivative of Multiple Terms in x

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

Our basic rule still applies when we have multiple terms in x.
We simply have to ensure that each individual term has been prepared
for differentiation. Then differentiate each term separately.

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

f ' ( x)  g ' ( x)  h' ( x)
then

Example 4
3
f ( x)  x  3x  9
4 2
1
Find the derivative of 2
2x
f ( x)  1 x 4  3 x 2  3 x 1  9
2 2
Solution:
f ' ( x)  4 ﾴ 1 x (41)  2 ﾴ 3x (21)  1ﾴ 3 x ( 11)  0
1. Prepare for differentiation 2 2
f ' ( x )  2 x 3  6 x  3 x 2
2
3
reduce power by 1” for each term f ' ( x)  2 x 3  6 x 
2x2
3. Tidy up

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

Example 5
Find the derivative of f(x) = (2x – 3) (3x + 6)

Solution: f ( x)  2 x(3 x  6)  3(3 x  6)
f ( x)  6 x 2  12 x  9 x  18
(ie expand brackets and simplify)
f ( x)  6 x 2  3 x  18
2. “Multiply by power then f ' ( x)  12 x  3

3. Tidy up

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

Example 6
x3  4 x 2  x  3 NAB
f ( x) 
Find the derivative of
x2
Solution:
x3 4 x 2 x 3
1. Re-write expression with each
f ( x)  2  2  2  2
term in the numerator over the x x x x
denominator
f ( x)  x  4  x 1  3 x 2
(ie use laws of indices)

3. “Multiply by power then f ' ( x)  1  x 2  6 x 3

16
f ( x)  1  2  3
'
4. Tidy up
x x

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

Leibniz Newton
Leibniz Notation
5.

dy  ax  bx  c
2
dx

Gottfried Wilhelm Leibnitz (1646 - 1716)
On the accession in 1714 of his master, George I.,
to the throne of England, Leibnitz was thrown aside
as 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 perhaps
might be expected of a professional diplomatist of that
time; but all who once came under the charm of his
personal presence remained sincerely attached to him.

He also held eminent positions in diplomacy, philosophy and literature.
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 .
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
Newton’s correspondence refers to “fluxions”
1666
and “the sum of infinitesimals”
1675 Leibniz makes the calculus public (published 1684)
1689 William of Orange takes English Crown
Massacre in Glencoe
1692
Newton publishes work on the calculus
1693
1694 Bank of England Formed
1695 Bank of Scotland Formed
1707 Act of Union
1712 First steam engine
George I begins Hanovarian dynasty leading
1714
to Robert Walpole becoming first Prime Minister.

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

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

If y is expressed in terms of x then the derivative is written as
/dx .
dy
“dee y by
dee x”
so /dx = 6x - 7 .
dy
eg y = 3x - 7x
2

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

Example 1 If Q = 9R2 - 15 find /dR !
dQ

R3
Solution:
Q  9 R 2  15 R 3

dQ
2. “Multiply by power then  2 ﾴ 9 R (21)  (3) ﾴ 15 R ( 31)
reduce power by 1” dR
dQ
 18 R  45 R 4
3. Tidy up
dR
dQ 45
 18R  4
dR R

Example 2 A curve has equation y = 5x3 - 4x2 + 7 .
Find the gradient of the tangent at x = -2
Solution:
y  5x  4 x  7
3 2
(NB. y =, so Leibniz notation)
dy
2. “Multiply by power then  15 x 2  8 x
dx

dy
 15(2) 2  8(2)
3. Substitute in given x-value
dx(2)
dy
 60  16
dx(2)
4. Communicate that you have Gradient at
dy
 76
answered question. x = -2 is 76
dx (2)

Extra bit for “FIZZY SYSTS” or even Physicists.

Newton’s 2ndLaw of Motion

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

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

ie speed or velocity
so /dt = u + at
ds

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

and this is Newton’s 1st Law of Motion

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

The Equation of the tangent
6.

dy
y  y1  m( x  x1 )
 f ( x)
'

dx

The Tangent to the Graph

y = mx + c y = f(x)

A (a,b)

tangent

So far we have differentiated functions, stating that the derived
function provides us with the gradient of the tangent to the graph.
We have seen that whilst the derived function is the same throughout
the curve the actual gradient is dependant on the x-coordinate.


y = mx + c y = f(x)

A (a,b)

tangent

Now we want to find the equation of the tangent.
The tangent is a straight line. We know from previous work on the
straight line that y  y1  m( x  x1 ) . How can we find m?


y = mx + c y = f(x)

A (a,b)

tangent

At any point gradient of curve = gradient of tangent
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 Copy the following:

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.

y  y1  m( x  x1 )
3. Substitute these details into:

Example 1 NAB
Find the equation of the tangent to the curve y  5 x  1
4
at x = 1

y  5x4  1
Solution:
y  5(1) 4  1 =6
1. If not given, find y-coordinate
dy
 20 x3
2. Find dy / dx
dx
dy
3. Find gradient at x-coordinate  20(1)3 = 20
dx(1)
Equation of tangent at (1,6):
y  6  20( x  1)
4. Substitute values into y  y1  m( x  x1 )
y  6  20 x  20
y  20 x  14

Example 2
Show that there is only one tangent to the curve y  5 x  6 x
2

with gradient 36
Solution:
1. Note we are being asked to prove
that dy/dx = 36 has only one
solution.
dy
 10 x  6
2. Find dy/dx dx
36  10 x  6
3. Make an equation with given
10 x  30
gradient and solve
x3
As dy/dx = 36 has only one
4. Make statement
solution there is only one
tangent with that gradient.

Example 3
Find the point of contact of the curve y  x  3 x  8
2
at which the
gradient is 9.
y  x 2  3x  8
Solution:
dy
1. Find dy / dx  2x  3
dx
2. Make dy / dx equal to given 2x  3  9
gradient and solve.
2x  6
x3
3. Substitute into original equation to y-coord when x = 3:
find y-coordinate.
y  (3) 2  3(3)  8
(DO NOT USE dy / dx.)
y  9  9  8  10

4. Make statement Point of contact (3 , 10)

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

Increasing / Decreasing
7.
Functions

f ( x) > 0 f ( x) < 0
'
'

Is the function increasing or decreasing?

y = mx + c y = f(x)
Whilst f(x) is increasing
gradient is POSITIVE.
A (a,b)

tangent

Our next task is to establish whether or not a function is increasing
or decreasing.
What can we say about the gradient of the tangent as the y- coordinates
increase i.e. the value of the function is increasing?


y = f(x)
Whilst f(x) is decreasing
gradient is NEGATIVE.

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


y = f(x)
Whilst f(x) is not changing
gradient is ZERO

What happens when the graph is changing direction?
The tangent is horizontal and so the gradient is ZERO.
These are called stationary points.

Increasing and Decreasing Functions Copy the following:

y = f(x)

f ' ( x) > 0
f ( x) < 0
'

f ( x)  0
'

f ( x) > 0
'
Function increasing
f ( x) < 0
'
Function decreasing
f ( x)  0
'
Stationary Point

Example 1

State whether the function y  4 x  3 x  3 is increasing or
3 2

decreasing at x = -2
y  4 x  3x  3
3 2

Solution:
dy
 12 x 2  6 x
1. Find dy / dx
dx
dy
 12 ﾴ (2) 2  6 ﾴ (2)
2. Find gradient at x-coordinate
dx(2)
dy
 48  12 = 36
dx(2)

3. Make statement As dy/dx is positive, function
is increasing at x = -2

Example 2
Find the intervals in which the function y  4 x  6 x  2
3 2

is increasing and decreasing.
y  4 x3  6 x 2  2
Solution:
dy
1. Find dy/dx  12 x 2  12 x
dx

2. Set dy/dx = 0 and solve to find 12 x 2  12 x  0
x-coords of Stationary Points
12 x( x  1)  0 Factorise
12 x  0 x 1  0
or

x0 x  1
or
3. Make a table to show what
-1 0
X
gradient is before and after SP’s
dy + - +
0 0
dx

Example 2
Find the interval in which the function y  4 x  6 x  2
3 2

is increasing and decreasing.
Solution:
x < 1
f(x) increasing:
4. Make statement
f(x) decreasing: 1 < x < 0

x>0
f(x) increasing:

-1 0
X
dy + - +
0 0
dx

The Proof
y  4 x3  6 x 2  2
x < 1
f(x) increasing:

f(x) decreasing: 1 < x < 0

x>0
f(x) increasing:

Example 3
x3 is never decreasing.
Show that the function y   3 x 2  9 x  4
3 x3
y   3x 2  9 x  4
Solution: 3
y  1 x3  3x 2  9 x  4
1. Prepare for differentiation 3

dy
 x2  6x  9
2. Find dy / dx dx
dy
 ( x  3) 2
3. Factorise (all terms involving x
dx
must have even powers).

Since (x - 3)2 will always
4. Make statement
produce a positive gradient
function is never decreasing.

The Proof
x3
y   3x 2  9 x  4
3

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

Stationary Points
8.

f ( x)  0
'

What if the function is neither increasing nor decreasing?

y = f(x)
Whilst f(x) is not changing
gradient is ZERO

Recall from the last lesson that when the gradient is changing there
will be points where the tangent is horizontal and so the gradient=ZERO.
At these points f ( x )  0
'
These are called stationary points.

The Nature of a Stationary Point Copy the following:

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

Rising Point of Inflection Falling Point of Inflection

Example 1 NAB
Determine the coordinates and nature of the stationary points on the
curve y  5 x ( x  4)
3

y  5 x3 ( x  4)
Solution:
y  5 x 4  20 x3
dy
 20 x 3  60 x 2
dx
2. Find dy / dx
At SP’s dy/dx = 0
3. Set dy/dx equal to zero, factorise 0  20 x3  60 x 2
if possible, and solve for x 0  20 x 2 ( x  3)
20 x 2  0 or ( x  3)  0
In exams must
x = 0 or x=3
make this
statement

Example 1

curve y  5 x ( x  4)
3

x=0 or x=3
Solution:
For x = 3:
4. Find y-coordinates by subbing these For x = 0:
values into original equation
y  5(0)3 ((0)  4) y  5(3) ((3)  4)
3
(NOT dy/dx)
y  5 ﾴ 27 ﾴ 1
y0
y  135

SP’s are (0,0) and (3,-135)
5. State coords of Stationary Points

Example 1

curve y  5 x ( x  4)
3

Solution: SP’s are (0,0) and (3,-135)
dy
6. Draw a NATURE TABLE for each  20 x 3  60 x 2
(0,0)
dx
SP to determine its nature.


0 0
0
X
dy
 20(1)3  60(1) 2  80 dy
d (1)
-ve 0 -ve
dx
dy
 20(1)3  60(1) 2  40
Slope
d (1)

7. Make statement (0,0) is a falling point of
inflexion.

Example 1

curve y  5 x ( x  4)
3

Solution: SP’s are (0,0) and (3,-135)
dy
6. Draw a NATURE TABLE for each  20 x 3  60 x 2
(3,-135)
dx
SP to determine its nature.


3 3
3
X
dy
 20(1)3  60(1) 2  40 dy
d (1)
-ve 0 +ve
dx
dy
 20(4)3  60(4) 2  320 Slope
d (4)

7. Make statement (3,-135) is a minimum
turning point.

The Proof
y  5 x3 ( x  4)

What do we need to know to sketch the curve?

Stationary Points
y = f(x)
y-intercept
x-intercept
Behaviour as x gets small

Behaviour as x gets big

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

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 )

xﾮﾥ
4. Behaviour as x gets very big and very small
x ﾮ ﾥ

Example 1

Sketch the graph of y  8 x  3 x
3 4

y  8 x3  3x 4
Solution:
y  8(0)3  3(0) 4 = 0
1. Find y-intercept (x = 0)
y-intercept is (0,0)

0  8 x3  3x 4
2. Find x-intercept (y = 0)
0  x3 (8  3 x)

or (8  3 x)  0
x3  0
3x  8
8
x
x=0
3
x-intercepts (0,0) & ( 2 2,0)
3

Example 1

3 4

y  8 x3  3x 4

dy
Solution:  24 x 2  12 x 3
dx
3. Find Stationary Points and their
At SP’s dy/dx = 0
nature (dy/dx = 0 )
0  24 x 2  12 x 3
0  12 x 2 (2  x)

12 x 2  0 or (2  x)  0
x=0 x =2
y  8(0)3  3(0) 4 y  8(2)3  3(2) 4
y = 16
y=0

Example 1
dy
3 4
 24 x 2  12 x 3
dx
SP’s are (0,0) and (2,16)
Solution:


2
0 0 2
X
3. Find Stationary Points and their
dy
nature (dy/dx = 0 ) -ve
0 0
+ve +ve
dx
dy Slo
 24(1) 2  12(1)3  36
d (1) pe
dy
 24(1) 2  12(1)3  12
d (1) (0,0) is a rising point of inflexion
dy (2,16) is a maximum tp
 24(3) 2  12(3)3  108
d (3)

Example 1 Biggest
power
3 4
dominates
Solution:
y  8 x3  3x 4
4. Behaviour as x gets very big and
very small x ﾮﾥ As x ﾮﾥ -3x will dominate
4

x ﾮ ﾥ and so large x’s will be -ve

Even Power - +ve
Odd Power – get what
As x ﾮ ﾥ -3x4 dominates
you started with
and so small x’s will be -ve

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

x-intercepts (0,0) & ( 2 2
Solution: y-intercept is (0,0) ,0)
3

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

(2,16)

(0,0)
22
3

x-intercepts (0,0) & ( 2 2
Solution: y-intercept is (0,0) ,0)
3

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

(2,16)

(0,0)
22
3

y  8 x3  3x 4

Heinemann , p.107, EX 6N
Q1, 2, 5, 7 & 8

Max /Min on Closed interval
10.

What is a closed interval?

y = f(x)
Closed interval

Up to now we have looked at graphs without any restrictions i.e we
have 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.

Where are the maximum and minimum values in a closed interval?
1 ﾣ x ﾣ 2
Lets consider f(x) = 8 + 2x – x2 on the interval
f(x)
x
Smallest
x f(x) x f(x) Largest 1 9
& SP
-1 5 0 8 1.1 8.99
-0.9 5.39 0.1 8.19 1.2 8.96
-0.8 5.76 0.2 8.36 1.3 8.91
-0.7 6.11 0.3 8.51 1.4 8.84
-0.6 6.44 0.4 8.64 1.5 8.75
-0.5 6.75 0.5 8.75 1.6 8.64
-0.4 7.04 0.6 8.84 1.7 8.51
-0.3 7.31 0.7 8.91 1.8 8.36
-0.2 7.56 0.8 8.96 1.9 8.19
-0.1 7.79 0.9 8.99 2 8

Maximum & Minimum Values on a Closed Interval
Copy the following:
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
In a closed interval the maximum and minimum values of a function
y
are either at

a stationary point or at
Local maximum
an end point of the interval.
x
Local minimum

Example 1

y  4  3x 2  x3
Find the maximum and minimum values of
within the interval 3 ﾣ x ﾣ 1
y  4  3x 2  x3
Solution: dy
 6 x  3 x 2
dx
1. Find Stationary Points
At SP’s dy/dx = 0

0  6 x  3 x 2
0  3x(2  x)
3 x  0 or (2  x)  0
x  2
x=0
y = 4 y  4  3(2)  (2)
2 3

y=0
SP’s are (0,4) & (-2,0)

Example 1

y  4  3x 2  x3
y  4  3x 2  x3
Solution: At x = -3
2. Find value of f(x) at each end point
y  4  3(3) 2  (3)3
of given interval
y  4  3(9)  (27)
y  4  27  27
(-3,4)
y=4

Example 1

y  4  3x 2  x3
y  4  3x 2  x3
Solution: At x = 1
2. Find value of f(x) at each end point
y  4  3(1) 2  (1)3
of given interval
y  4  3(1)  (1)
y  4  3 1
(1,0)
y=0

3. Compare Y-COORDINATES to (0,4) & (-2,0) (-3,4) (1,0)
find smallest and biggest

Maximum = 4, Minimum = 0
4. Make statement

Heinemann , p.109, EX 6O
Q2 (a), (b), (e) & (f)

Graph of the Derived
11.
Function

What will the graph of a derivative look like?
y = x3 – 6x2 + 9x

Gradient –ve
Gradient
to SP 2
+ve
Gradient +ve
to SP 1

Imagine you were asked to sketch a graph of f ' ( x)
What would the key points be?

Is the gradient constant? y = x3 – 6x2 + 9x
Lets consider dy/dx = 3x2 – 12x + 9
dy/dx
x
x dy/dx x dy/dx
3.25 1.688
-1 24 1.25 -1.31
3.5 3.75
-0.75 19.7 1.5 -2.25 Min
Moving 3.75 6.188
-0.5 15.8 to zero = -3
1.75 -2.81
4 9
-0.25 12.2 2 -3
4.25 12.19
0 9 2.25 -2.81
4.5 15.75
SP at
0.25 6.19 2.5 -2.25
(3,0) 4.75 19.69
0.5 3.75 SP at 2.75 -1.31
5 24
(1,0)
0.75 1.69 3 0
1 0
Decreasing to (2,-3) then increasing

The original and the derived
y  f ' ( x) y = f(x)

Example 1
Sketch the derived function for f(x)
y = f(x)

x
SP occurs 5
at x = 5
- 0 + New y-values
f '(x)

Example 1

Negative but
approaching zero

x
SP occurs 5
at x = 5
- 0 + New y-values
f '(x)

Example 1
y  f ' ( x)

Positives

Negative but
approaching zero

x
SP occurs 5
at x = 5
- 0 + New y-values
f '(x)

Example 2
y = f(x)

(2,-4)

x 0 2
SPs occur at
f'(x) + 0 - 0 + New y-values
x=0 &x=2

Example 2

(2,-4)

x 0 2
SPs occur at
x=0 &x=2

Example 2

Positive but
approaching zero

(2,-4)

x 0 2
SPs occur at
x=0 &x=2

Example 2

Goes into –ve’s
but returns to zero by 2 (2,-4)

x 0 2
SPs occur at
x=0 &x=2

Example 2

Into +ve’s

x 0 2
SPs occur at
x=0 &x=2

Optimisation : Maxima &
12.
Minima

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.

Optimisation
Optimisation 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.

Optimisation
Problems 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.

Copy the following:
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.

Example 1 (Formula Given)
The “Smelly Place” Garden Centre has a model train which is used
to 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:
400
C  50  25V 
V
where V is the speed in miles per hour.

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

400
C  50  25V 
Example 1
V
Solution:

C  50  25V  400V 1
1. Find derivative, remembering
to prepare for differentiation.
dC
 25  400V 2
dV
dC 400
 25  2
dV V

400
C  50  25V 
Example 1
V
Solution:
dC 400
 25  2
dV V

2. Make statement then At SP’s dC/dV = 0
set derivative equal to zero.
400
0  25  2
V
400 25 Cross Multiply

V2 1
400  25V 2 Reject
–ve speed
V 2  16
V 4

400
C  50  25V 
Example 1
V
Solution:
dC 400
 25  2 V 4
dV V
3. Justify nature of each SP
using a nature table.
4
X 4 4
dC 400
 25  2  375 dC
-ve 0
d (1) (1) +ve
dV
dC 400 Slope
 25  2  17
d (5) (5)

A minimum tp occurs and so cost
4. Make statement
is minimised at V = 4.

400
C  50  25V 
Example 1
V
A minimum tp occurs and so cost
Solution:
is minimised at V = 4.

400
4. Sub x-value found into C  50  25V 
V
original expression to find
corresponding minimum value. For V = 4:
(DO NOT USE dy/dx)
400
C  50  25(4) 
4
C  50  100  100
C = £250
Minimum cost is £250 per month
5. Answer Question
and occurs when speed = 4mph

Heinemann , p.115, EX 6R
Q3

Example 2 (Perimeter and area)
A desk is designed which is rectangular in shape and which is required
in 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
y

x
Solution to (a):
Area = length x breadth
1. Start with what you know

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

2. In order to differentiate we
Perimeter = 2x + 2y
need expression only involve
X’s. Use other information to
420  2 x  2 y
find an expression for Y.
Divide
2 y  420  2 x
by 2
3. Change subject to y.
y  210  x

Example 2
y  210  x
Area = x x y
Solution to (a):

4. Now substitute this expression Area = x x (210 – x)
for y into our original expression
for AREA. Area  A( x)  210 x  x 2
Solution to (b):
Aﾢ x)  210  2 x
(
5. To find dimensions giving
maximum or minimum first find At SP’s A’(x) = 0
derivative.
210  2 x  0
6. Make statement and set
2 x  210
derivative equal to zero.
x  105

Example 2
y  210  x
Area = x x y
Solution to (a):
x  105
7. Now use your expression for y y  210  105  105
to find other dimension.

Aﾢ x)  210  2 x
Example 2 (
X = 105 and Y = 105
Solution:

8. Must justify maximum
values using nature table.
105
X
Aﾢ
(100)  210  2(100)  10
Aﾢ x) +ve 0
( -ve
Aﾢ
(110)  210  2(110)  10
Slope

The area is maximised when
9. Make statement
length is 105cm and breadth is
105 cm.

Example 2 The area is maximised when
length is 105cm and breadth is
105 cm.
Solution to (c):
Area  A( x)  210 x  x 2
10. To find actual maximum area
sub maximum x-value just
Areamax  210(105)  (105) 2
found into expression for area.
(DO NOT USE dy/dx)
Areamax  22, 050  11, 025

Areamax  11, 025 cm 2

PROOF

Heinemann , p.112, EX 6Q
Q1, 2, 5

Example 3 (Surface Area and Volume)
A cuboid of volume 51.2 cm3 is made with a length 4 times its
breadth.
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 area
and calculate this area.

xcm

Example 3
h
x
Solution to (a):
4x
1. Start with what you know Surface Area = sum of area of faces
= 2(4x x h) + 2(x x h)+ 2(4x x x)
Lets call height h, so:
= 10xh + 8x2
2. In order to differentiate
Volume = 4x x x x h
expression must only involve
X’s. Use other information to
ﾴ
51.2  4x 2 h
find an expression for h.

51.2 12.8
h 2
3. Change subject to h. 2
4x x

Surface Area = sum of area of faces
Example 2
= 10xh + 8x2
12.8
Solution to (a):
h 2
x
4. Now substitute this expression
for y into our original expression
￦ 12.8
( )
A( x)  10 ﾴ x ﾴ 2  8 x 2
for SURFACE AREA.
x
￨
￦x
128
( )
A( x)   8x2
x2
￨

128
A( x)  8 x 
2

x

128
Example 3
A( x)  8 x 
2

x
Solution to (b): A( x)  8 x 2  128 x 1
5. To find dimensions giving
128
Aﾢ x)  16 x  2
maximum or minimum first find (
x
derivative.
At SP’s A’(x) = 0
6. Make a statement and set 128 Multiply
16 x  2  0
derivative equal to zero. by x2
x
16 x 3  128  0
16 x 3  128
x3  8
x 382

12.8
Example 3
h 2
x
Solution to (b):
12.8 12.8
h   3.2
2
(2) 4
7. To find corresponding height
substitute x value just found into
So dimensions minimising area:
expression for h found in step 2.

Breadth = 2 cm

Length = 8 cm

Height = 3.2 cm

128
Example 3
Aﾢ x)  16 x  2
(
x
x 382
Solution:

8. Must justify minimum
values using nature table.
X 2
Aﾢ  16  128  112
(1)
Aﾢ x) -ve 0
( +ve
128
Aﾢ  16(4)   56
(4)
16 Slope

The surface area is minimised when
9. Make statement
the breadth is 2 cm

The surface area is minimised when
Example 3
the breadth is 2 cm
128
Solution to (c): A( x)  8 x 
2

x
10. To find actual maximum area
128
Areamin  8(2) 
2
sub maximum x-value just
2
found into expression for area.
(DO NOT USE dy/dx)
Areamin  32  64

Areamin  96 cm 2

PROOF

Heinemann , p.112, EX 6R
Q1, 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 placed
Into the channel to protect the cables.
100 cm

x
4
0

c
x
m

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 V ( x)  200 x(20  x)
(b) Calculate the value of x which gives the maximum volume of gutter
and find this volume.

Example 2 (a) Show that the volume of each section of guttering is
V ( x)  200 x(20  x)
Solution to part (a):
Require volume so: Length = 100 cm
V  l ﾴ bﾴ h Breadth = (40 – 2x) cm

V  100 ﾴ (40  2 x) ﾴ x Height = “folded bit” = x cm

V  100 x(40  2 x) Factorise
bracket
V  100 x ﾴ 2(20  x)
V  200 x(20  x) as required.

(b) Calculate the value of x which gives the maximum
Example 2
volume of gutter. V ( x)  200 x(20  x)
Maximum and
Solution to part (b):
minimum will
V ( x)  200 x(20  x) occur at SP’s

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

V ' ( x)  4000  400 x
At SP’s V ' ( x)  0 Must make this statement

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

Example 2

x = 10 Now prove this is a maximum tp

V ' ( x)  4000  400 x
 
X 10 10
10
dV
-ve
+ve 0
dx
Slope

So when x = 10 we have a maximum

Example 2

So when x = 10 we have a maximum

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

V (10)  200 ﾴ 10 ﾴ (20  10)

V (10)  20000cm3

So maximum volume of guttering Make statement
is 20,000 cm3.

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