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

Section 3.4 The Chain Rule 203
ExamplE 9 Differentiate y − e sec 3 .
SOLUTION The outer function is the exponential function, the middle function is the
secant function, and the inner function is the tripling function. So we have
dy
sec 3
d
− e ssec 3d
d d
− e sec 3 sec 3 tan 3 d
d s3d
− 3e sec 3 sec 3 tan 3
■
How to Prove the Chain Rule
Recall that if y − f sxd and x changes from a to a 1 Dx, we define the increment of y as
Dy − f sa 1 Dxd 2 f sad
According to the definition of a derivative, we have
Dy
lim
Dx l 0 Dx − f 9sad
So if we denote by « the difference between the difference quotient and the derivative,
we obtain
lim « − lim
Dx l 0 Dx l 0S D Dy
Dx 2 f 9sad − f 9sad 2 f 9sad − 0
But
« − Dy 2 f 9sad ? Dy − f 9sad Dx 1 « Dx
Dx
If we define « to be 0 when Dx − 0, then « becomes a continuous function of Dx. Thus,
for a differentiable function f, we can write
7 Dy − f 9sad Dx 1 « Dx where « l 0 as Dx l 0
and « is a continuous function of Dx. This property of differentiable functions is what
enables us to prove the Chain Rule.
Proof of the Chain Rule Suppose u − tsxd is differentiable at a and y − f sud is differentiable
at b − tsad. If Dx is an increment in x and Du and Dy are the corresponding
increments in u and y, then we can use Equation 7 to write
8 Du − t9sad Dx 1 « 1 Dx − ft9sad 1 « 1 g Dx
where « 1 l 0 as Dx l 0. Similarly
9 Dy − f 9sbd Du 1 « 2 Du − f f 9sbd 1 « 2 g Du
where « 2 l 0 as Du l 0. If we now substitute the expression for Du from Equation 8
into Equation 9, we get
Dy − f f 9sbd 1 « 2 gft9sad 1 « 1 g Dx
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