11DIFFERENTIATION - Department of Mathematics
11DIFFERENTIATION - Department of Mathematics
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)