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.
Let’s recall the rules on indices …..

Rule a 0 = 1 a -m =
1 a m n a m  a m n a m × a n = a m+n a m
 a n = a m-n
(a m ) n = a mn
Examples
12.314
0 =
1 x -5 = x
5
1
1 n
3
 n

3
3 x 2  x
2
3
2a 3/2 × 3a 1/2 = 6
3 1 a a 2
= 6 a x 2
 x -3 = x x
2

3
(q 2 ) 3 =
5
= q
2   q
2
2
= q
6 y 5
6
= 5 y
6

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.

The 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
B(x
2, y
2
)
A(x
1
,y
1
) m x
2
– x
1
 y
2
– y
1
The gradient of AB = m
AB
C y
2
 y
1 x
2
 x
1
, Gradient Formula

Change in y
Change in x
 y
2
 y
1 x
2
 x
1

We will investigate the tangent to a curve at a single point.
The points A and B lie on a function. The line joining A and
B is not a tangent to the curve at the point A. If we think about moving the point B towards A, the slope of the line will change.
We can observe the results by watching this graphically. We will take the simple function f(x) = x 2
Click the graphic to investigate this function.

Observe the table of results and suggest a relationship between the value of the slope of the tangent and the value of x on the function
f(x) = x 2
It appears that the relationship may be f ′(x) = 2x
Can we find this algebraically?
1
0 x
2
-1
-2 f ′ (x)
4
2
0
-2
-4

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 at a point on a curve.
f(x)
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)) 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.
f(x) x x+h x

The gradient of the line AB would therefore be m
AB
 change in y change in x m m
AB

AB
 y
2
y
1 x
2
x
1

 x f(x) m
AB
 x+h f(x+h)
(
 
( ) h

Look at what happens as we move
B towards A, i.e. h is getting smaller.
As the size of h gets smaller and smaller the line AB becomes the tangent to the curve at the point
A. We refer to this as the limit as h → 0. Note that h cannot become 0 or we would not get a line AB!
The gradient of the line AB becomes the same as the gradient of the curve at A
This is the value we were looking for.

Differentiate the function f(x) = x 2 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 lim h

0

( )
 lim h

0
(
  h f x
 lim h

0
( x
 h )
2  x
2 h
( ) f x
 lim h

0 x
2 
2 xh h xh
 h
2 f x
 lim h

0
2 h h (2 x
 h ) f x
 lim h

0 h f x
 lim(2 h

0 x
 h )

( )

2 x
 h
2  x
2 h gets so small its effectively zero.

2
4
4
For each and every point on the curve f(x)=x 2 , the gradient of the tangent to the curve at the point x is given by the formula f ′(x)=2x
10
10 m
= -6
8
8
6
6
A table of results might make this clearer m
= -4 m =
-2
4
4
2
2 m = 0 m =
2
2
2 m
= 4
4
4
Value of x Gradient of the tangent
-3
-6
-2 -4
-1
0
1
2
3
4
-2
4
6
8
0
2
We can agree the observation graphically and
Is there an easier way to do this?
algebraically.

f(x) f ′ (x) f(x) f ′ (x) f(x) f ′ (x) x 4 x 5 x 2 x 3 x n
2x
3x 2
4x 3
5x 4 nx n-1
2x 2
3x 2
4x 2
5x 2 ax 2
4x
6x
8x
10x
2ax
3x 3 3
3
5x 4 4
4
7x 3 3 3
6x 7 7 7 ax n
9x 2
20x 3
21x 2
42x 6 anx n-1

There are four rules for differentiating – remember these and you can differentiate anything …
Rule f(x) = x n
 f ′(x) = nx n-1 f(x) = cx n
 f ′ (x) = cnx n-1 f(x) = c
 f ′ (x) = 0
Examples f(x) = x 6
 f ′ (x) = 6 x 6 - 1 = 6 x 5 f(x)= 4x 2
 f ′ (x) = 4 x
2 x 2-1 f(x) = 65
 f ′ (x) =
= 8 x 1 or 8 x
0 f(x) = g(x) + h(x)
 f ′ (x) = g′ (x) + h′ (x) f(x)= x 6 + 4x 2 + 65
 f ′ (x) = 6 x 5 + 8x

There are many different ways of writing f ′ (x): f ′ (x) dy dx df dx d dx y ′ y ′ (x)
The most common of these are: f ′ (x)  functional notation used when function is defined as f(x) dy dx
 Leibnitz notation used when function is defined as y =

The Gradient of a Tangent
Remember:
Differentiation is used to find the gradient of a tangent to a graph.
Find the equation of the tangent to the curve y = 3x 3 – x + 6 at x = 2 .
A tangent is a straight line, so we need to use: y – b = m(x – a)
To use this we need to know:
 the gradient of the tangent at the point on the curve
(in this case when x = 2)
 the coordinates of a point on the line (in this case (2, ?) )

The Gradient of a Tangent
Step 1 : Finding the gradient when x = 2
Differentiate the function y

3 x
3  
6
As the function is given in terms of y we will use the dy / dx notation y

3 x
3  
6 dy dx

9 x
2 
1
At the point x = 2, dy dx

9(2)
2 
1 dy dx

35 So the gradient of the tangent is 35

Step 2 :
The Gradient of a Tangent
Finding the coordinates at the point
At the point x = 2 the function value is given by
y = 3 (2) 3 – 2 + 6
y = 3(8) + 4
y = 28
The coordinate is therefore the point (2, 28)

Step 3 :
The Gradient of a Tangent
Finding the equation of the tangent
The line passing through (2,28) with gradient 35 is :
 
(

) y

28

35( x

2) y

28

35 x

70 y

35 x

42
The equation of the tangent at x = 2 is y – 35x + 42 = 0

You can confirm this by checking on a graphing calculator or by sketch.
– 3 – 2 – 1
– 20
– 40
– 60
100 y
80
60
40
20
1 2 3 4 5 x

Sketching Graphs of Derived Functions
We can investigate graphs of derived functions by looking at a dynamic setting of a function, along with the resulting derived function.
There are 3 examples shown on separate pages.
Look at each one and build up the derived graph as you are prompted. Try to predict the derivative of the third graph before it is drawn.
Click the graphic to start your investigation

Sketching Graphs of Derived Functions
We can sketch the graph of the derived function, f ′(x), by considering the graph of f(x).
The diagram shows the graph of y = f(x)
Sketch the graph of y = f ′ (x)
Step 1 Decide what type of function it is: (-1,4) y y = f(x)
This looks like a cubic function, i.e. it has an x 3 term in it.
(3,-2) x
When we differentiate the function the x 3 term will become an x 2 term.
 f ′ (x) will be a quadratic function and will look like this or

Sketching Graphs of Derived Functions
Step 2 Consider the gradient of the function at key points
Tangent is horizontal m = 0 y Gradient is negative m < 0 y = f(x)
Gradient is positive m > 0
(-1,4) x
(3,-2)
Gradient is positive m > 0
Tangent is horizontal m = 0
This can be summarised on a table

Sketching Graphs of Derived Functions y y = f(x)
(-1,4) x
(3,-2) x → f ′ (x) + ve
-1
0
→
- ve
3
0
→
+ ve

Sketching Graphs of Derived Functions
Step 3 x f ′ (x)
Sketch this information as a graph
→
+ ve 0
→
- ve 0
→
+ ve x x f’(x) +ve above the line f’(x)
= 0 on the line f’(x)
-ve below the line