University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 175: 73


$$y'=\frac{1}{\theta\ln 2}$$

Work Step by Step

$$y=\log_2(5\theta)$$ Recall Theorem 7: $$\frac{d}{d\theta}\log_au=\frac{1}{u\ln a}\frac{du}{d\theta}$$ Apply here to find $y'$: $$y'=(\log_2(5\theta))'=\frac{1}{5\theta\ln 2}(5\theta)'=\frac{5}{5\theta\ln 2}$$ $$y'=\frac{1}{\theta\ln 2}$$
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.