Calculus 10th Edition

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



Work Step by Step

$y=\dfrac{1}{(x-2)}=(x-2)^{-1}.$ Using the Chain Rule: $u=(x-2)$; $\dfrac{du}{dx}=1$ $\dfrac{dy}{du}=-\dfrac{1}{u^2}$ $\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}=-\dfrac{1}{(x-2)^2}.$
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