Higher Mathematics:
Unit 1.3
Introduction to Differentiation
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.
Let’s recall the rules on indices …..
Rules of indices
Rule
12.3140 = 1
a0 = 1
1
a-m =
a
n
m
a a
m
Examples
m
n
1
x5
x
-5 =
3
x 
2
x
1

3
n
2
3
6
5
y =
3
2
am
×
an
=
am+n
2a3/2 × 3a1/2 = 6a a
= 6a 2
23
am  an = am-n
(am)n = amn
x2 x -3 = x
5
x
=
2
2
2
2
3
q

q

q
(q ) =
= q6
1
2
n
5
3
y6
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 curves
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(x2,y2)
y2 – y1
A(x1,y1)
The gradient of AB = mAB
x 2 – x1
C
y2  y1
m
, Gradient Formula
x2  x1
Change in y

Change in x
y2  y1

x 2  x1
Tangents to curves
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) = x2
Click the graphic to investigate this function.
Tangents to curves
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) = x2
x
f ′ (x)
2
4
1
2
It appears that the relationship
may be
f ′(x) = 2x
0
0
-1
-2
-2
-4
Can we find this algebraically?
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 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
x+h
Tangents to curves
The gradient of the line AB would
therefore be
change in y
mAB 
change in x
y2 - y1
mAB 
x2 - x1

mAB 

x+h f(x+h)
x f(x)
mAB
f ( x  h)  f ( x )

h
Tangents to curves
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.
Putting it together….
Differentiate the function f(x) =
from first principles.
x2
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.
f ( x)  lim
h 0
( x  h) 2  x 2
f ( x)  lim
h 0
h
x 2  2 xh  h 2  x 2
f ( x)  lim
h 0
h
2 xh  h 2
f ( x)  lim
h 0
h
h(2 x  h)
f ( x)  lim
h 0
h
h gets so
f ( x)  lim(2 x  h)
small its
h 0
The limit as h → 0 is
written as lim
h0
f ( x  h)  f ( x )
h
f ( x)  2 x
effectively
zero.
What does the answer mean?
For each and every point on the
curve f(x)=x2 , the gradient of
the tangent to the curve at
the point x is given by the
formula
f ′(x)=2x
y
10
10
10
y
y
m=
8
8
8
-6
-4
– 4
–
– 4
4
-2
-1
3
m
=
– 2
–
– 2
2
-6
-4
2
=4
m
4
4
4
m= 0 m
-3
1
6
6
6
– 2
–
– 2
2
Gradient of the tangent
0
2might make this
A table of results
2
2
m
=
clearer
=2
-2
– 4
–
– 4
4
Value of x
2
2
2
4
4
4
x
x
x
4
-2
0
2
4
6
8
Is there an easier way to do
We can agree
the
– 6
– 6
6
–
this?
– 8 graphically and
observation
– 8
8
–
– 10
–
algebraically.
– 10
10
Spot the pattern
f(x)
f(x)
ff ′′ (x)
(x)
f(x)
f(x)
fff ′′′ (x)
(x)
(x)
f(x)
f(x)
f(x)
fff′′′(x)
(x)
(x)
x22
2x
2x222
2x
4x
3x333
3x
3x
9x2
x3
3x2
3x22
3x
6x
5x444
5x
5x
20x3
x4
4x3
4x22
4x
8x
7x333
7x
7x
21x2
x5
5x4
5x22
5x
10x
6x777
6x
6x
42x6
xn
nxn-1
ax
ax22
2ax
ax
ax
axnnn
anxn-1
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) =
cxn
 f ′ (x) =
cnxn-1
f(x) = x6  f ′ (x) = 6 x 6 - 1 = 6 x 5
f(x)= 4x2  f ′ (x) = 4 x 2 x 2-1
= 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)
Notation
There are many different ways of writing f ′ (x):
f ′ (x)
dy
dx
df
dx
d
dx
( f ( x))
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 = 3x3 – 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  3x3  x  6
As the function is given in terms of y we will use the dy/dx notation
y  3x3  x  6
dy
dx
 9x  1
2
At the point x = 2,
dy
dx
 9(2)  1
dy
dx
 35
2
So the gradient of the tangent is 35
The Gradient of a Tangent
Step 2 :
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)
The Gradient of a Tangent
Step 3 :
Finding the equation of the tangent
The line passing through (2,28) with gradient 35 is :
y  b  m( x  a )
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
The Gradient of a Tangent
100
20
40
60
80
– 60
20
40
1– 3
2
3
4
5
1
2
You can confirm this
by checking on a
graphing calculator
or by sketch.
y
100
80
60
40
20
– 3
– 2
– 1
1
– 20
– 40
– 60
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:
y
y = f(x)
(-1,4)
This looks like a cubic function,
i.e. it has an x3 term in it.
x
(3,-2)
When we differentiate the function
the x3 term will become an x2 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
y
Tangent is
horizontal
m=0
Gradient is
negative
m<0
Gradient is
positive
m>0
y = f(x)
(-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)
→
-1
+ ve
0
→
3
→
- ve
0
+ ve
Sketching Graphs of Derived Functions
Step 3
x
f ′ (x)
Sketch this information as a graph
→
-1
→
3
3
→
+ ve
0
- ve
0
+ ve
(x)
ff '' (x)
f’(x) +ve above
the line
xx
f’(x) = 0 on
the line
f’(x) -ve below
the line