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 174: 12



Work Step by Step

$$y=\frac{1}{\ln3x}$$ Recall the Derivative Rule for Natural Logarithm: $$\frac{d}{dx}(\ln u)=\frac{1}{u}\frac{du}{dx}$$ We have $$y'=\Big(\frac{1}{\ln3x}\Big)'=\frac{(1)'\ln3x-1(\ln3x)'}{\ln^23x}=\frac{0\times\ln3x-\frac{1}{3x}(3x)'}{\ln^23x}$$ $$y'=\frac{0-\frac{3}{3x}}{\ln^23x}=-\frac{\frac{1}{x}}{\ln^23x}=-\frac{1}{x\ln^23x}$$
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