Differential calculus is concerned with the rate at which things change.
For example, the speed of a car is the rate at which the distance it travels changes with time.
First we shall review the gradient of a straight line graph, which represents a rate of change.
Gradient of a straight line graph
The gradient of the line between two points (x1, y1) and (x2, y2) is
change in y  y2  y1  m
change in x x2  x1
where m is a fixed number called a constant.
A gradient can be thought of as the rate of change of y with respect to x.
Gradient of a curve
A curve does not have a constant gradient. Its direction is continuously changing, so its
gradient will continuously change too.
y = f(x)
The gradient of a curve at any point on the curve is defined as being the gradient of
the tangent to the curve at this point.
A tangent is a straight line, which touches, but does not cut, the curve.
y
A
O
Tangent to
the curve
at A.
x
We cannot calculate the gradient of a tangent directly, as we know only one point
on the tangent and we require two points to calculate the gradient of a line.
Using geometry to approximate to a gradient
Look at this curve.
y
B1
B2
B3
Tangent to
the curve
at A.
A
O
x
Look at the chords AB1, AB2, AB3, . . .
For points B1, B2, B3, . . . that are closer and closer to A the sequence of chords
AB1, AB2, AB3, . . . move closer to becoming the tangent at A.
The gradients of the chords AB1, AB2, AB3, . . . move closer to becoming the gradient
of the tangent at A.
A numerical approach to rates of change
Here is how the idea can be applied to a real example. Look at the section of the
graph of y = x2 for 2 > x > 3. We want to find the gradient of the curve at A(2, 4).
x
changes
from
y
changes
from
gradient
AB1
2 to 3
4 to 9
94
=5
3 2
AB2
2 to 2.5
4 to 6.25
6.25  4
= 4.5
2.5  2
AB3
2 to 2.1
4 to 4.41
4.41  4
= 4.1
2.1  2
AB4
2 to 2.001
4 to 4.004001
AB5
2 to
2.00001
4 to
4.0000400001
B1 (3, 9)
Chord
B2 (2.5, 6.25)
B3 (2.1, 4.41)
B4 (2.001, 4.004001)
Complete
the table
A (2, 4)
The gradient of the chord AB1 is
94  5
3 2
4.004001  4
= 4.001
2.001  2
4.00001
As the points B1, B2, B3, . . . get closer and closer to A the gradient is getting closer to 4.
This suggests that the gradient of the curve y = x2 at the point (2, 4) is 4.
y = x2
y
4
2
x
Example (1)
Find the gradient of the chord joining the two points with x-coordinates 1 and 1.001 on
the graph of y = x2. Make a guess about the gradient of the tangent at the point x = 1.
The gradient of the chord is
1.0012 1  1.0020011
1.0011
1.0011
 0.002001
0.001
(1.001, 1.0012)
= 2.001
I’d guess 2.
(1, 1))
Example (2)
Find the gradient of the chord joining the two points with x-coordinates 8 and 8.0001
on the graph of y = x2. Make a guess about the gradient of the tangent at the point x = 8.
The gradient of the chord is
8.00012  64  64.00160001 64
8.0001 8
8.00018
 0.00160001
0.0001
(8.0001, 8.00012 )
= 16.0001
I’d guess 16.
(8, 64)
Let’s make a table of the results so far:
You’re probably noticing a pattern
here. But can we prove it
mathematically?
x-coordinate
gradient
1
2
2
4
3
4
5
6
7
8
16
I need to consider what happens
when I increase x by a general
increment. I will call it h.
I will call it ∆x.
(2 + h, (2 + h)2)
(2, 4)
h
Let y = x2 and let A be the point (2, 4)
Let B be the point (2 + h, (2 + h)2)
y
Here we have increased x by a very small
amount h. In the early days of calculus h
was referred to as an infinitesimal.
y = x2
B(2 + h, (2 + h)2)
Draw the chord AB.
A(2, 4)
2
2
(
2

h
)

4
4

4
h

h
4

Gradient of AB 
2 h2
2 h2
 4h  h
h
 h( 4  h)
h
=4+h
2
O
x
If h ≠ 0 we can
cancel the h’s.
As h approaches zero, 4 + h approaches 4.
So the gradient of the curve at the point (2, 4) is 4.
Use a similar method to find the
gradient of y = x2 at the points
(i) (3, 9)
(ii) (4, 16)
We can now add to our table:
It looks like the gradient is
simply 2x.
x-coordinate
gradient
1
2
2
4
3
6
4
8
5
6
7
8
16
Let’s check this result.
y
y = x2
15
10
5
x
−4
−3
−2
−1
1
2
3
4
Let’s check this result.
y
y = x2
15
10
Gradient at
(3, 9) = 6
5
x
−4
−3
−2
−1
1
2
3
4
Let’s check this result.
y
y = x2
15
10
Gradient at
(2, 4) = 4
5
x
−4
−3
−2
−1
1
2
3
4
Let’s check this result.
y
y = x2
15
10
Gradient at
(1, 1) = 2
5
x
−4
−3
−2
−1
1
2
3
4
Let’s check this result.
y
y = x2
15
10
Gradient at
(0, 0) = 0
5
x
−4
−3
−2
−1
1
2
3
4
Let’s check this result.
y
y = x2
15
10
Gradient at
(–1, 1) = –2
5
x
−4
−3
−2
−1
1
2
3
4
Let’s check this result.
y
y = x2
15
10
Gradient at
(–2, 4) = –4
5
x
−4
−3
−2
−1
1
2
3
4
Let’s check this result.
y
y = x2
15
10
Gradient at
(–3, 9) = –6
5
x
−4
−3
−2
−1
1
2
3
4
Another way of seeing what the gradient is at the point (2, 4) is to plot an accurate graph
and ‘zoom in’.
9
y
y = x2
8
7
6
ZOOM IN
5
4
3
2
1
x
−4
−2
2
4
5
y
4.5
When we zoom in the curve
starts to look like a straight
line which makes it easy to
estimate the gradient.
0.8
4
Gradient  0.8  4
0.2
0.2
3.5
x
3
1
1.5
2
2.5
3