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: 74



Work Step by Step

$$y=\log_3(1+\theta\ln3)$$ Recall Theorem 7: $$\frac{d}{d\theta}\log_au=\frac{1}{u\ln a}\frac{du}{d\theta}$$ Apply here to find $y'$: $$y'=(\log_3(1+\theta\ln3))'=\frac{1}{(1+\theta\ln3)\ln3}(1+\theta\ln3)'$$ $$y'=\frac{(0+1\times\ln3)}{(1+\theta\ln3)\ln3}=\frac{\ln3}{(1+\theta\ln3)\ln3}$$ $$y'=\frac{1}{1+\theta\ln3}$$
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