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11DIFFERENTIATION - Department of Mathematics

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722 11 DIFFERENTIATION<br />

FIGURE 11.5<br />

The composite function h(x) g[ f(x)]<br />

Rule 7: The Chain Rule<br />

the function f, we might suspect that the rule h(x) for the derivative h <strong>of</strong> h<br />

will be given by an expression that involves the rules for the derivatives <strong>of</strong> f<br />

and g. But how do we combine these derivatives to yield h?<br />

This question can be answered by interpreting the derivative <strong>of</strong> each<br />

function as giving the rate <strong>of</strong> change <strong>of</strong> that function. For example, suppose<br />

u f(x) changes three times as fast as x—that is,<br />

f(x) du<br />

3<br />

dx<br />

And suppose y g(u) changes twice as fast as u—that is,<br />

g(u) dy<br />

2<br />

du<br />

Then, we would expect y h(x) to change six times as fast as x—that is,<br />

h(x) g(u) f(x) (2)(3) 6<br />

or equivalently,<br />

dy dy<br />

<br />

dx du<br />

du<br />

(2)(3) 6<br />

dx<br />

This observation suggests the following result, which we state without pro<strong>of</strong>.<br />

If h(x) g[ f(x)], then<br />

h(x) d<br />

g( f(x)) g( f(x)) f(x) (1)<br />

dx<br />

Equivalently, if we write y h(x) g(u), where u f(x), then<br />

REMARKS<br />

h = g[ f ]<br />

f g<br />

x f (x) = u y = g(u) = g[ f (x)]<br />

dy dy du<br />

<br />

dx du dx<br />

1. If we label the composite function h in the following manner<br />

Inside function<br />

<br />

h(x) g[ f(x)]<br />

<br />

Outside function<br />

then h(x) is just the derivative <strong>of</strong> the ‘‘outside function’’ evaluated at the<br />

‘‘inside function’’ times the derivative <strong>of</strong> the ‘‘inside function.’’<br />

(2)

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