James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

202 Chapter 3 Differentiation Rules
ExamplE 7 Differentiate y − e sin x .
More generally, the Chain Rule gives
d
dx seu d − e du u dx
SOLUTION Here the inner function is tsxd − sin x and the outer function is the exponential
function f sxd − e x . So, by the Chain Rule,
dy
dx − d
dx se sin x d − e d sin x dx ssin xd − e sin x cos x
We can use the Chain Rule to differentiate an exponential function with any base
b . 0. Recall from Section 1.5 that b − e ln b . So
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and the Chain Rule gives
b x − se ln b d x sln bdx
− e
d
dx sb x d − d
dx se sln bdx sln bdx
d
d − e sln bdx
dx
− e sln bdx ∙ ln b − b x ln b
Don’t confuse Formula 5 (where x is
the exponent) with the Power Rule
(where x is the base):
d
dx sx n d − nx n21
because ln b is a constant. So we have the formula
5
In particular, if b − 2, we get
d
dx sb x d − b x ln b
6
In Section 3.1 we gave the estimate
d
dx s2x d − 2 x ln 2
d
dx s2x d < s0.69d2 x
This is consistent with the exact formula (6) because ln 2 < 0.693147.
The reason for the name “Chain Rule” becomes clear when we make a longer chain
by adding another link. Suppose that y − f sud, u − tsxd, and x − hstd, where f , t, and
h are differentiable functions. Then, to compute the derivative of y with respect to t, we
use the Chain Rule twice:
dy
dt
− dy
dx
ExamplE 8 If f sxd − sinscosstan xdd, then
dx
dt − dy
du
du
dx
dx
dt
f 9sxd − cosscosstan xdd d dx
cosstan xd
− cosscosstan xddf2sinstan xdg d stan xd
dx
− 2cosscosstan xdd sinstan xd sec 2 x
Notice that we used the Chain Rule twice.
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