Cambridge International AS and A Level Mathematics Pure Mathematics 1

Differentiation

P1^

5

EXAMPLE 5.1 Find the gradient of the curve y = x^3 at the general point (x, y).

SOLUTION

Let P have the general value x as its x co-ordinate, so P is the point (x, x^3 ) (since

it is on the curve y = x^3 ). Let the x co-ordinate of Q be (x + h) so Q is

((x +h), (x + h)^3 ). The gradient of the chord PQ is given by

QR

PR

= +

+

= +++

=

+

() –

()–

–

xh x

xh x

xxhxhh x

h

xh

33

32 23 3

2

33

3 33

33

33

23

2 2

2 2

xh h

h

hx xh h

h

xxhh

+

= ++

=+ +

( )

As Q takes values closer to P, h takes smaller and smaller values and the gradient

approaches the value of 3x^2 which is the gradient of the tangent at P. The

gradient of the curve y = x^3 at the point (x, y) is equal to 3x^2.

Note

If the equation of the curve is written as y = f(x), then the gradient function (i.e. the

gradient at the general point (x, y)) is written as f’(x). Using this notation the result

above can be written as f(x) = x^3 ⇒ f’(x) = 3 x^2.

x

y

P

Q

R

O

Figure 5.7