Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.5 The Chain Rule - 2.5 Exercises: 53

Answer

$$y=-x+\pi$$

Work Step by Step

First we will find the derivative: $$y'=(\sin(\sin x))'=\cos(\sin x)(\sin x)'=\cos(\sin x)\cos x$$ The tangent line of the curve at the point $(x_{0},y_{0})$ is given by formula: $$y-y_{0}=y'(x_{0})(x-x_{0}).$$ For $(x_{0},y_{0})=(\pi,0)$ we get: $$y-0=\left.\cos(\sin x_{0})\cos x_{0}\right|_{x_{0}=\pi}(x-\pi)\Rightarrow y=\cos(\sin \pi)\cos\pi(x-\pi)\Rightarrow y=\cos0(-1)(x-\pi)\Rightarrow y=1\cdot(-1)(x-\pi)\Rightarrow y=-x+\pi$$
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