Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.3 - Exponential Functions - Exercises 7.3 - Page 391: 68

Answer

$$\frac{{dy}}{{d\theta }} = \frac{1}{{1 + \theta \ln 3}}$$

Work Step by Step

$$\eqalign{ & y = {\log _3}\left( {1 + \theta \ln 3} \right) \cr & {\text{Find the derivative of }}y{\text{ with respect to }}\theta \cr & \frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {{{\log }_3}\left( {1 + \theta \ln 3} \right)} \right] \cr & {\text{use the rule }}\frac{d}{{dx}}\left[ {{{\log }_a}u} \right] = \frac{1}{{\ln a}} \cdot \frac{1}{u}\frac{{du}}{{dx}}. \cr & \frac{{dy}}{{d\theta }} = \frac{1}{{\ln 3}} \cdot \frac{1}{{1 + \theta \ln 3}}\frac{d}{{d\theta }}\left[ {1 + \theta \ln 3} \right] \cr & {\text{solve the derivative}} \cr & \frac{{dy}}{{d\theta }} = \frac{1}{{\ln 3}} \cdot \frac{1}{{1 + \theta \ln 3}}\left( {\ln 3} \right) \cr & {\text{simplifying}} \cr & \frac{{dy}}{{d\theta }} = \frac{1}{{1 + \theta \ln 3}} \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.