## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 13 - The Trigonometric Functions - Chapter Review - Review Exercises - Page 701: 64

#### Answer

$$\frac{{dy}}{{dx}} = - \tan x$$

#### Work Step by Step

\eqalign{ & y = \ln \left| {\cos x} \right| \cr & {\text{differentiate with respect to }}x \cr & \frac{{dy}}{{dx}} = {D_x}\left( {\ln \left| {\cos x} \right|} \right) \cr & {\text{use the chain rule }}{D_x}\left( {\ln u} \right) = \frac{1}{u} \cdot {D_x}\left( u \right).{\text{ consider }}u = \cos x \cr & \frac{{dy}}{{dx}} = \left( {\frac{1}{{\cos x}}} \right){D_x}\left( {\cos x} \right) \cr & {\text{solve the derivative }} \cr & \frac{{dy}}{{dx}} = \left( {\frac{1}{{\cos x}}} \right)\left( { - \sin x} \right) \cr & {\text{then}} \cr & \frac{{dy}}{{dx}} = - \frac{{\sin x}}{{\cos x}} \cr & {\text{simplifying}} \cr & \frac{{dy}}{{dx}} = - \tan x \cr}

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