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: 56



Work Step by Step

$$y=\ln(3\theta e^{-\theta})$$ We have $$(\ln \theta)'=\frac{1}{\theta}$$ $$(e^{\theta})'=e^\theta$$ So the derivative of $y$ is $$y'=\frac{1}{3\theta e^{-\theta}}(3\theta e^{-\theta})'=\frac{3(e^{-\theta}+\theta e^{-\theta}(-\theta)')}{3\theta e^{-\theta}}$$ $$y'=\frac{e^{-\theta}-\theta e^{-\theta}}{\theta e^{-\theta}}=\frac{e^{-\theta}(1-\theta)}{\theta e^{-\theta}}$$ $$y'=\frac{1-\theta}{\theta}$$
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