## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 3 - Derivatives - 3.9 Derivatives of Logarithmic and Exponential Functions - 3.9 Exercises: 57

#### Answer

${y^,} = - \sin x\,\left( {2+\ln \,\left( {{{\cos }^2}x} \right)} \right)$

#### Work Step by Step

$\begin{gathered} y = \cos x\ln \,\left( {{{\cos }^2}x} \right) \hfill \\ \hfill \\ Differentiate,\,{\text{ using}}\,\,{\text{the product rule}} \hfill \\ \hfill \\ {\text{and}}{\left( {{\text{cos}}\left( u \right)} \right)^,} = - \sin u \cdot {u^,} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ {y^,} = \cos x\,\left( {\frac{{2\cos x\,\left( { - \sin x} \right)}}{{{{\cos }^2}x}}} \right) + \ln \,\left( {{{\cos }^2}x} \right)\,\left( { - \sin x} \right) \hfill \\ \hfill \\ multiply \hfill \\ \hfill \\ {y^,} = - 2\sin x - \sin x\ln \,\left( {{{\cos }^2}x} \right) \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ {y^,} = - \sin x\,\left( {2+\ln \,\left( {{{\cos }^2}x} \right)} \right) \hfill \\ \end{gathered}$

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