Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.5 Exercises: 27

Answer

$\dfrac{dy}{dx}=\dfrac{1-\sec^2{(x+y)}}{\sec^2{(x+y)}}.$ At $(0, 0)\rightarrow\dfrac{dy}{dx}=\dfrac{1-\sec^2{(0+0)}}{\sec^2{(0+0)}}=0.$

Work Step by Step

$\dfrac{d}{dx}(\tan{(x+y)})=\dfrac{d}{dx}(x)\rightarrow$ Using the Chain Rule with $u=x+y\rightarrow\dfrac{du}{dx}=1+\dfrac{dy}{dx}\rightarrow$ $\dfrac{d}{dx}(\tan{(x+y)})=(1+\dfrac{dy}{dx})\sec^2{(x+y)}\rightarrow$ $(1+\dfrac{dy}{dx})\sec^2{(x+y)}=1\rightarrow$ $\dfrac{dy}{dx}=\dfrac{1-\sec^2{(x+y)}}{\sec^2{(x+y)}}.$ At $(0, 0)\rightarrow\dfrac{dy}{dx}=\dfrac{1-\sec^2{(0+0)}}{\sec^2{(0+0)}}=0.$
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