## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 7: Transcendental Functions - Section 7.2 - Natural Logarithms - Exercises 7.2 - Page 381: 25

#### Answer

$$\frac{{dy}}{{d\theta }} = 2\cos \left( {\ln \theta } \right)$$

#### Work Step by Step

\eqalign{ & y = \theta \left( {\sin \left( {\ln \theta } \right) + \cos \left( {\ln \theta } \right)} \right) \cr & {\text{Find the derivative of }}y{\text{ with respect to }}\theta \cr & \frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {\theta \left( {\sin \left( {\ln \theta } \right) + \cos \left( {\ln \theta } \right)} \right)} \right] \cr & {\text{use the product rule}} \cr & \frac{{dy}}{{d\theta }} = \theta \frac{d}{{d\theta }}\left[ {\sin \left( {\ln \theta } \right) + \cos \left( {\ln \theta } \right)} \right] + \left( {\sin \left( {\ln \theta } \right) + \cos \left( {\ln \theta } \right)} \right)\frac{d}{{d\theta }}\left[ \theta \right] \cr & {\text{use the rules }}\cr & \frac{d}{{d\theta }}\left[ {\sin u} \right] = \cos u\frac{{du}}{{d\theta }}{\text{ and }}\frac{d}{{d\theta }}\left[ {\cos \theta } \right] = - \sin \theta \frac{{du}}{{d\theta }} \cr & \frac{{dy}}{{d\theta }} = \theta \left( {\cos \left( {\ln \theta } \right)\frac{d}{{d\theta }}\left[ {\ln \theta } \right] - \sin \left( {\ln \theta } \right)\frac{d}{{d\theta }}\left[ {\ln \theta } \right]} \right) + \left( {\sin \left( {\ln \theta } \right) + \cos \left( {\ln \theta } \right)} \right)\frac{d}{{d\theta }}\left[ \theta \right] \cr & {\text{solve the derivatives}} \cr & \frac{{dy}}{{d\theta }} = \theta \left( {\cos \left( {\ln \theta } \right)\left( {\frac{1}{\theta }} \right) - \sin \left( {\ln \theta } \right)\left( {\frac{1}{\theta }} \right)} \right) + \left( {\sin \left( {\ln \theta } \right) + \cos \left( {\ln \theta } \right)} \right)\left( 1 \right) \cr & {\text{simplifying, we get:}} \cr & \frac{{dy}}{{d\theta }} = \cos \left( {\ln \theta } \right) - \sin \left( {\ln \theta } \right) + \sin \left( {\ln \theta } \right) + \cos \left( {\ln \theta } \right) \cr & \frac{{dy}}{{d\theta }} = 2\cos \left( {\ln \theta } \right) \cr}

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