Cambridge International AS and A Level Mathematics Pure Mathematics 1

Differentiation

P1^

5

Using differentiation

EXAMPLE 5.5 Given that yx

x

=−^82 , find

(i) d

d

y

x

(ii) the gradient of the curve at the point (4, 1^12 ).

SOLUTION

(i) Rewrite yx

x

=−^82 as yx=− x−

(^12)
8 2.
Now you can differentiate using the rule yk ==xn⇒ ddyx knxn−^1.
d
d
y
x
xx
x x

=+

=+

1 − −

2

(^12)
16
1
2

16

3

3

(ii) At (4, 1^12 ), x = 4

Substituting x = 4 into the expression for d

d

y

x

gives

(^) d
d
y
x

=+

=+

=

1

24

16

43

1

4

16

64

1

2

EXAMPLE 5.6 Figure 5.10 shows the graph of

y = x^2 (x − 6) = x^3 − 6 x^2.

Find the gradient of the curve at the points A and B where it meets the x axis.

$ %

y y = x (^3) –x 2
x
Figure 5.10