## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 13 - The Trigonometric Functions - 13.2 Derivatives of Trigonometric Functions - 13.2 Exercises - Page 688: 18

#### Answer

$$\frac{{dy}}{{dx}} = - \frac{{3\sin \left( {\ln \left| {2{x^3}} \right|} \right)}}{x}$$

#### Work Step by Step

\eqalign{ & y = \cos \left( {\ln \left| {2{x^3}} \right|} \right) \cr & {\text{differentiate with respect to }}x \cr & \frac{{dy}}{{dx}} = {D_x}\left[ {\cos \left( {\ln \left| {2{x^3}} \right|} \right)} \right] \cr & {\text{use the chain rule }}{D_x}\left( {\cos u} \right) = - \sin u \cdot {D_x}\left( u \right).{\text{ consider }}u = \ln \left| {2{x^3}} \right| \cr & \frac{{dy}}{{dx}} = - \sin \left( {\ln \left| {2{x^3}} \right|} \right) \cdot {D_x}\left( {\ln \left| {2{x^3}} \right|} \right) \cr & {\text{use the chain rule }}{D_x}\left( {\ln u} \right) = \frac{1}{u} \cdot {D_x}\left( u \right) \cr & \frac{{dy}}{{dx}} = - \sin \left( {\ln \left| {2{x^3}} \right|} \right)\left( {\frac{1}{{2{x^3}}}} \right){D_x}\left( {2{x^3}} \right) \cr & {\text{solve the derivative and simplify}} \cr & \frac{{dy}}{{dx}} = - \sin \left( {\ln \left| {2{x^3}} \right|} \right)\left( {\frac{1}{{2{x^3}}}} \right)\left( {6{x^2}} \right) \cr & \frac{{dy}}{{dx}} = - \sin \left( {\ln \left| {2{x^3}} \right|} \right)\left( {\frac{3}{x}} \right) \cr & \frac{{dy}}{{dx}} = - \frac{{3\sin \left( {\ln \left| {2{x^3}} \right|} \right)}}{x} \cr}

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