Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 45

Answer

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

Work Step by Step

Given $y=\cos{\sqrt{\sin(\tan{\pi x})}}$. $y'=\frac{d\cos{\sqrt{\sin(\tan{\pi x})}}}{dx}$ Using the chain rule: $y'$ $=\frac{d\cos{\sqrt{\sin(\tan{\pi x})}}}{d\sqrt{\sin(\tan{\pi x})}} \times \frac{d\sqrt{\sin(\tan{\pi x})}}{d\sin(\tan{\pi x})} \times \frac{d\sin(\tan{\pi x})}{d\tan{\pi x}} \times \frac{{d\tan{\pi x}}}{d\pi x} \times \frac{d\pi x}{dx}$ $=-\sin{\sqrt{\sin(\tan{\pi x})}} \times \frac{1}{2\sqrt{\sin(\tan{\pi x})}} \times \cos(\tan{\pi x}) \times \sec^2{\pi x} \times \pi$ Simplifying the above expression, $y' =-\frac{1}{2}\pi\frac{\sin{\sqrt{\sin(\tan{\pi x})}} \times \cos(\tan{\pi x}) \times \sec^2{\pi x}}{\sqrt{\sin(\tan{\pi x})}}$
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