## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 6 - Applications of Integration - 6.8 Logarithmic and Exponential - 6.8 Exercises: 10

#### Answer

$$- \frac{{2\sin x}}{{\cos x}}$$

#### Work Step by Step

\eqalign{ & \frac{d}{{dx}}\left( {\ln \left( {{{\cos }^2}x} \right)} \right) \cr & {\text{use the rule }}\frac{d}{{dx}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dx}},{\text{ letting }}u = {\cos ^2}x \cr & = \frac{1}{{{{\cos }^2}x}}\frac{d}{{dx}}\left( {{{\cos }^2}x} \right) \cr & {\text{by the chain rule}} \cr & = \frac{1}{{{{\cos }^2}x}}\left( {2\cos x} \right)\frac{d}{{dx}}\left( {\cos x} \right) \cr & = \frac{1}{{{{\cos }^2}x}}\left( {2\cos x} \right)\left( { - \sin x} \right) \cr & {\text{simplify}} \cr & = - \frac{{2\sin x}}{{\cos x}} \cr}

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