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: 55

Answer

$$y'=-2\tan\theta$$

Work Step by Step

$$y=\ln(\cos^2\theta)$$ The derivative of $y$ is $$y'=\Big(\ln(\cos^2\theta)\Big)'=\frac{1}{\cos^2\theta}\times(\cos^2\theta)'$$ $$y'=\frac{1}{\cos^2\theta}\times2\cos\theta\times(\cos\theta)'=\frac{2}{\cos\theta}\times(-\sin\theta)$$ $$y'=-\frac{2\sin\theta}{\cos\theta}=-2\tan\theta$$
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