The Chemistry Maths Book, Second Edition

106 Chapter 4Differentiation

can be differentiated by first expanding the cube and then differentiating term by

term:

y 1 = 1 (2x

2

1 − 1 1)

3

1 = 18 x

6

1 − 112 x

4

1 + 16 x

2

1 − 11

A simpler way is to use the chain rule. Letu 1 = 12 x

2

1 − 11 and rewrite yas a function

of u:

y 1 = 1 g(u) 1 = 1 u

3

Then by Rule 5 in Table 4.3,

and, substituting for u,

In this example, yhas been considered in two ways:

(i) as a function of x:f(x) 1 = 1 (2x

2

1 − 1 1)

3

;

(ii) as a function of u:g(u) 1 = 1 u

3

where uis the functionu 1 = 12 x

2

1 − 11. The substitution

u 1 = 12 x

2

1 − 11 highlights the structure of the function, that of a cube, and makes the

chain rule the natural method of differentiation.

EXAMPLE 4.14Differentiatey 1 = 1 (2x

2

1 − 1 1)

522

.

Puty 1 = 1 u

522

, whereu 1 = 12 x

2

1 − 11. Then

0 Exercises 37– 40

Table 4.4 shows the generalization of Table 4.2 for elementary functions of a function,

f(u)whereu 1 = 1 u(x).

dy

dx

dy

du

du

dx

=×=uxxx













=−

5

2

41021

32 2 32

() ( )

dy

dx

=−12 2 1xx

22

()

dy

dx

dy

du

du

dx

=×= ×()()34ux

2

dy

dx

=−+=48 48 12 12 2 1xxxxx−

53 22

()