Differentiation

2. Introduction
• You will learn what Differentiation is
• You will see how to apply it to solve
graph based problems
• It is one of the single most important
topics there is in Maths!

3. Differentiation
You need to be able to find the gradient
function of a formula
On a straight line graph, the gradient is
constant, the same everywhere along the
line.
On any curved graph, the gradient is always
changing. Its value depends on where you are
along the x-axis.
The different gradients can be shown by
tangents. These are lines that touch the
curve in only one place.
Differentiation is the process whereby we
can find a formula to give the gradient on a
curve, at any point on it.
7B
Straight Line =
Constant
Gradient
Curved line =
Gradient
Changes
tangent
tangent

4. Differentiation
You need to be able to find the gradient
function of a formula
As a general rule, if;
then…
7B
( ) n
f x ax

1
'( ) n
f x nax 

This is the
formula for the
curve, ie) the
function
This is the gradient
function for f(x).
The formula that
gives the gradient at
any point.
Its is also know as
the derivative, or
‘derived function’
Examples
a) f(x) = x3
Find the derived function of each of
the following…
f’(x) = 3x2
b) f(x) = 2x2
f’(x) = 4x1 (4x)
c) f(x) = x-2
f’(x) = -2x-3
d) f(x) = -3x-3
f’(x) = 9x-4

5. Differentiation
You need to be able to find the gradient
function of a formula
As a general rule, if;
then…
7B
( ) n
f x ax

1
'( ) n
f x nax 

This is the
formula for the
curve, ie) the
function
This is the gradient
function for f(x).
The formula that
gives the gradient at
any point.
Its is also know as
the derivative, or
‘derived function’
Examples
e)
Find the derived function of each of
the following…
2
1
( )
f x
x

2
( )
f x x

3
'( ) 2
f x x
 
Must be written in
the form axn first!
f) ( )
f x x

1
2
( )
f x x

1
2
1
'( )
2
f x x


1
'( )
2
f x
x

Must be written in
the form axn first!

6. Differentiation
You need to be able to solve Graphical
problems using the Gradient Function
Remember that differentiating gives us a
formula for the gradient at a given point on
the graph (x).
A standard question will ask you to work out
the gradient of a curve at a particular point.
This is when differentiating is used.
7C/D
Examples
a) Calculate the gradient of the curve
f(x) = x2 where x = 3
2
( )
f x x

'( ) 2
f x x

'(3) 6
f 
Differentiate to get
the gradient function
Substitute in the
value for x at the
required point
b) Calculate the gradient of the curve
f(x) = x2 at the coordinate (-2,4)
2
( )
f x x

'( ) 2
f x x

'( 2) 4
f   
Differentiate to get
the gradient function
Substitute in the
value for x at the
required point
y = x2

7. Differentiation
You need to be able to solve Graphical
problems using the Gradient Function
Remember that differentiating gives us a
formula for the gradient at a given point on
the graph (x).
A standard question will ask you to work out
the gradient of a curve at a particular point.
This is when differentiating is used.
Examples
c) Find dy/dx when y equals x2 – 6x - 4
2
6 4
y x x
  
dy
dx
 2x 6

Differentiate
each term
separately.
A number on
its own
disappears
This also means the
gradient function
7C/D

8. Differentiation
You need to be able to solve Graphical
problems using the Gradient Function
Remember that differentiating gives us a
formula for the gradient at a given point on
the graph (x).
A standard question will ask you to work out
the gradient of a curve at a particular point.
This is when differentiating is used.
Examples
d) Find dy/dx when y equals 3 – 5x2.
2
3 5
y x
 
dy
dx
 10x

Differentiate
each term
separately.
A number on
its own
disappears
7C/D

9. Differentiation
You need to be able to solve Graphical
problems using the Gradient Function
Remember that differentiating gives us a
formula for the gradient at a given point on
the graph (x).
A standard question will ask you to work out
the gradient of a curve at a particular point.
This is when differentiating is used.
Examples
e) Let f(x) = 4x2 – 8x + 3. Find the
gradient of the curve y = f(x) at the
point (1/2, 0)
2
4 8 3
y x x
   Differentiate to
get the gradient
function
8 8
dy
x
dx
 
   
1 1
2 2
' 8 8
f   
 
1
2
' 4
f  
Substitute in
x = 1/2
7C/D

10. Differentiation
You need to be able to solve Graphical
problems using the Gradient Function
Remember that differentiating gives us a
formula for the gradient at a given point on
the graph (x).
A standard question will ask you to work out
the gradient of a curve at a particular point.
This is when differentiating is used.
Examples
f) Find the coordinates when the graph
y = 2x2 - 5x + 3 has a gradient of 7.
2
2 5 3
y x x
   Differentiate to
get the gradient
function
'( ) 4 5
f x x
 
The gradient is 7 at the point we want
7 4 5
x
 
12 4x

3 x

Add 5
Divide by 4
The x coordinate where the gradient is
7 has a value of 3. Substitute this into
the ORIGINAL function to find the y-
coordinate
2
2(3) 5(3) 3
y   
2
2 5 3
y x x
  
6
y 
So the graph has a gradient of 7 at
(3,6)
7C/D

11. Differentiation
You need to be able to solve Graphical
problems using the Gradient Function
Remember that differentiating gives us a
formula for the gradient at a given point on
the graph (x).
A standard question will ask you to work out
the gradient of a curve at a particular point.
This is when differentiating is used.
Examples
g) Find the gradient of the curve y = x2
where it meets the line y = 4x - 3
2
4 3
x x
 
Set the equations equal to each other (to
represent where they meet)
2
4 3 0
x x
  
( 3)( 1) 0
x x
  
3 or 1
x x
 
Group on
the left
Factorise
The lines will meet at (3,9) and
(1,1) by substitution. Differentiate y = x2 to get the gradient
function
2
y x

2
dy
x
dx

At (3,9) the gradient will be 6
(by putting ‘3’ into the gradient
function)
At (1,1) the gradient will be 2
(by putting ‘1’ into the gradient
function)
7C/D

12. Differentiation
You need to be able to deal with much
more complicated equations when
differentiating
Remember as before, all terms must be
written in the form axn before they can be
differentiated.
It is useful to note that at this stage, being
able to rewrite in this way is not essential.
However being able to switch between forms
will be very useful on harder questions.
Being able to do this now is worth practising
as you will definitely need it on C2/3/4!
7E
Examples
Differentiate the following:
a)  
3
3 1
x x 
Rewrite in the
form axn
4 3
3x x

dy
dx
 3 2
12 3
x x

dy
dx
  
2
3 4 1
x x 
Differentiate
Factorise

13. Differentiation
You need to be able to deal with much
more complicated equations when
differentiating
Remember as before, all terms must be
written in the form axn before they can be
differentiated.
It is useful to note that at this stage, being
able to rewrite in this way is not essential.
However being able to switch between forms
will be very useful on harder questions.
Being able to do this now is worth practising
as you will definitely need it on C2/3/4!
7E
Examples
Differentiate the following:
b)
1
x
1
2
x

3
2
1
2
dy
x
dx

 
dy
dx

1
2
 
3
2
x

dy
dx

1
2
  3
1
x
dy
dx

3
1
2 x

Rewrite in the
form axn
Differentiate
Imagine the term
was split apart
Rewrite the x
term using power
rules
Group the
fractions by
multiplying
tops/bottoms

14. Differentiation
You need to be able to deal with much
more complicated equations when
differentiating
Remember as before, all terms must be
written in the form axn before they can be
differentiated.
It is useful to note that at this stage, being
able to rewrite in this way is not essential.
However being able to switch between forms
will be very useful on harder questions.
Being able to do this now is worth practising
as you will definitely need it on C2/3/4!
7E
Examples
Differentiate the following:
c) 2
2
x
x

Split into 2
fractions
2 2
2
x
x x

2
1 2
x x

1 2
2
x x
 

2
x
 3
4x

dy
dx

2
1
x
 3
4
x

dy
dx

3
x
x
 3
4
x

dy
dx
  3
4
x
x
 
Cancel x’s on the
first one
Rewrite in the
form axn
Differentiate
Rewrite using
power rules
Make the
Denominators
common

15. Differentiation
You can repeat the process of
differentiation to get the ‘second
order derivative’
7F
( )
f x
'( )
f x
dy
dx
''( )
f x
2
2
d y
dx
or
or
Original Equation
Differentiate
once (first order
derivative)
Differentiate
twice (second
order derivative)
Examples
2
2
Find and of the following:
dy d y
dx dx
5
2
4
) 3
a y x
x
 
5
2
4
( ) 3
f x x
x
 
5 2
( ) 3 4
f x x x
 
4 3
'( ) 15 8
f x x x
 
3 4
''( ) 60 24
f x x x
 
Rewrite in the
form axn
Differentiate
Differentiate
again

16. Differentiation
You can repeat the process of
differentiation to get the ‘second
order derivative’
7F
( )
f x
'( )
f x
dy
dx
''( )
f x
2
2
d y
dx
or
or
Original Equation
Differentiate
once (first order
derivative)
Differentiate
twice (second
order derivative)
Examples
2
2
Find and of the following:
dy d y
dx dx
1
) 3
2
b y x
x
 
1
( ) 3
2
f x x
x
 
1
1
2
1
( ) 3
2
f x x x
 
1
2
3
'( )
2
f x x


3
2
3
''( )
4
f x x

 
Rewrite in the
form axn
Differentiate
Differentiate
again
2
1
2
x

3
x


17. Differentiation
You can use differentiation to find
the tangent to a curve at a
particular point, as well as the
normal at that point
Remember the curve we have is
based on an equation
The tangent is a straight line that
intersects the curve at on point
only.
 The gradient of the tangent is the
same as the gradient of the curve at
the point given (so you can
differentiate to get it)
The ‘normal’ is a straight line
perpendicular to the tangent where
is touches the curve.
7H
Curve
Tangent
Normal

18. Differentiation
You can use differentiation to find
the tangent to a curve at a particular
point, as well as the normal at that
point
Find the equation of the tangent to the
curve y = x3 – 3x2 + 2x - 1, at the point
(3,5).
So we need the gradient of the tangent.
It will be the same as the gradient of
the curve at the point given (x = 3)
7H
3 2
3 2 1
y x x x
   
2
3 6 2
dy
x x
dx
  
2
(3) 3(3) 6(3) 2
f   
(3) 11
f 
Differentiate to
get the gradient
function
Substitute the x
value in to find
the gradient
The gradient at
(3,5) is 11
1 1
( )
y y m x x
  
5 11( 3)
y x
  
5 11 33
y x
  
11 28
y x
 
Substitute the
co-ordinate and
gradient in
Expand bracket
Add 5
Use the formula for a straight line!
Curve
Tangent

19. Differentiation
You can use differentiation to find
the tangent to a curve at a particular
point, as well as the normal at that
point
Find the equation of the normal to the
curve y = 8 - 3√x at the point where x =
4.
Start by finding the gradient of the
curve at that point.
Gradient = -3/4 (this is of the tangent)
The Normal is perpendicular to the
tangent
Gradient of the Normal = 4/3
7H
8 3
y x
 
1
2
8 3
y x
 
1
2
3
2
dy
x
dx

 
3 1
( )
2
f x
x
  
3 1
(4)
2 4
f   
3
(4)
4
f  
Rewrite
Differentiate
Rewrite for
substitution
Sub in x = 4
The Gradient
where x = 4 is -3/4

20. Differentiation
You can use differentiation to find
the tangent to a curve at a particular
point, as well as the normal at that
point
Find the equation of the normal to the
curve y = 8 - 3√x at the point where x =
4.
Gradient of the Normal = 4/3
Now we need the co-ordinates at that
point. We already have x = 4.
Co-ordinates at the intersection = (4,2)
7H
8 3
y x
 
Substitute in x = 4
8 3 4
y  
2
y 
3 times 2 = 6!
1 1
( )
y y m x x
  
4
2 ( 4)
3
y x
  
Substitute the
co-ordinate and
gradient in
Multiply by 3
Expand bracket
Use the formula for a straight line!
3 6 4( 4)
y x
  
3 6 4 16
y x
  
3 4 10 0
y x
  
Rearrange

21. Differentiation
Showing how differentiation
works (sort of!)
y = x2
(x, x2)
(x+δx, (x+δx)2)
This is a lowercase ‘delta’,
representing a small increase
in x
x+δx - x
(x+δx)2 – x2
Gradient = change in
y ÷ change in x
Multiply the
bracket
Group some
terms
Cancel δx
At the original
point, δx = 0

22. Differentiation
Showing how differentiation
works (sort of!)
y = f(x)
(x, f(x))
(x+δx, f(x+δx))
This is a lowercase ‘delta’,
representing a small increase
in x
x+δx - x
f(x+δx) – f(x)
Gradient = change in
y ÷ change in x
Simplify the
denominator
This is the ‘definition’ for
differentiating a function
(you won’t really need this
specifically until C3)