Answer
$y'=-\dfrac{1}{(x-2)^2}.$
Work Step by Step
$y=\dfrac{1}{(x-2)}=(x-2)^{-1}.$
Using the Chain Rule:
$u=(x-2)$; $\dfrac{du}{dx}=1$
$\dfrac{dy}{du}=-\dfrac{1}{u^2}$
$\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}=-\dfrac{1}{(x-2)^2}.$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 2 - Differentiation - 2.4 Exercises - Page 136: 17
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