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: 31

Answer

$$y'=-4\sin(\sec4x)\sec4x\tan4x$$

Work Step by Step

$$y'=(\cos(\sec4x))'=-\sin(\sec4x)(\sec4x)'= -\sin(\sec4x)(\frac{1}{\cos 4x})'=-\sin(\sec4x)(-\frac{1}{\cos^24x})(\cos4x)'=\sin(\sec4x)\frac{1}{\cos^24x}(-\sin4x)(4x)'= -\sin(\sec4x)\frac{1}{\cos4x}\frac{\sin4x}{\cos4x}\cdot4= -4\sin(\sec4x)\sec4x\tan4x$$
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