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: 27

Answer

$$y'=-\frac{1}{2}\sqrt{\sin x+1}$$

Work Step by Step

$$y'=\left(\frac{\cos x}{\sqrt{1+\sin x}}\right)'=\frac{(\cos x)'\sqrt{1+\sin x}-\cos x(\sqrt{1+\sin x})'}{(\sqrt{1+\sin x})^2}=\frac{-\sin x(\sqrt{1+\sin x})-\cos x\frac{1}{2\sqrt{1+\sin x}}(1+\sin x)'}{1+\sin x}= \frac{-2\sin x(1+\sin x)-\cos x\cdot\cos x}{2(1+\sin x)^{3/2}}=\frac{-2\sin x-\sin^2x-(\sin^2x+\cos^2x)}{2(1+\sin x)^{3/2}}=-\frac{\sin^2x+2\sin x+1}{2(1+\sin x)^{3/2}}=-\frac{(\sin x+1)^2}{2(\sqrt{1+\sin x})^{3/2}}=-\frac{1}{2}\sqrt{\sin x+1}$$
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