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: 13



Work Step by Step

$$y=\ln(t^2)$$ Recall the Derivative Rule for Natural Logarithm: $$\frac{d}{dx}(\ln u)=\frac{1}{u}\frac{du}{dx}$$ Therefore, we have $$y'=\frac{1}{t^2}(t^2)'=\frac{2t}{t^2}=\frac{2}{t}$$
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