## Calculus 8th Edition

Published by Cengage

# Chapter 2 - Derivatives - 2.5 The Chain Rule - 2.5 Exercises - Page 158: 45

#### Answer

$$y'=\frac{-\pi\sin\sqrt{\sin(\tan \pi x)}\cos(\tan \pi x)\sec^2\pi x}{2\sqrt{\sin(\tan \pi x)}}$$

#### Work Step by Step

$$y'= \left(\cos\sqrt{\sin(\tan \pi x)}\right)'=-\sin\sqrt{\sin(\tan \pi x)}\cdot\left(\sqrt{\sin(\tan \pi x)}\right)'=-\sin\sqrt{\sin(\tan \pi x)}\frac{1}{2\sqrt{\sin(\tan \pi x)}}\cdot(\sin(\tan \pi x))'=\frac{-\sin\sqrt{\sin(\tan \pi x)}}{2\sqrt{\sin(\tan \pi x)}}\cdot \cos(\tan \pi x)\cdot(\tan \pi x)'= \frac{-\sin\sqrt{\sin(\tan \pi x)}\cos(\tan \pi x)}{2\sqrt{\sin(\tan \pi x)}}\frac{1}{\cos^2(\pi x)}(\pi x)'= \frac{-\sin\sqrt{\sin(\tan \pi x)}\cos(\tan \pi x)}{2\sqrt{\sin(\tan \pi x)}}\sec^2\pi x\cdot \pi= \frac{-\pi\sin\sqrt{\sin(\tan \pi x)}\cos(\tan \pi x)\sec^2\pi x}{2\sqrt{\sin(\tan \pi x)}}$$

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