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
()