Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.6 - Derivatives of Logarithmic Functions - 3.6 Exercises: 7

Answer

$$f'(x)=\frac{-\sin x}{\ln {10}(1+\cos x)}$$

Work Step by Step

$f'(x)=\frac{d}{dx}\log_{10}(1+\cos x)$ $=\frac{d}{dx}\frac{\ln(1+\cos x)}{\ln 10}$ $=\frac{1}{\ln 10}\frac{d}{dx}\ln (1+\cos x)$ Using the chain rule: $f'(x)=\frac{1}{\ln 10}[\frac{d\ln(1+\cos x)}{d1+\cos x} \times \frac{d1+\cos x}{dx}]$ $=\frac{1}{\ln 10}[\frac{1}{1+\cos x}\times-\sin x]$ $=\frac{-\sin x}{\ln {10}(1+\cos x)}$
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