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: 32



Work Step by Step

$$y=\ln(\sec\theta+\tan\theta)$$ Recall the following Derivative Rules: $$\frac{d}{d\theta}(\ln u)=\frac{1}{u}\frac{du}{d\theta}$$ $$\frac{d}{d\theta}(u+v)=\frac{du}{d\theta}+\frac{dv}{d\theta}$$ We have $$y'=\Big(\ln(\sec\theta+\tan\theta)\Big)'$$ $$y'=\frac{1}{\sec\theta+\tan\theta}(\sec\theta+\tan\theta)'=\frac{\sec\theta\tan\theta+\sec^2\theta}{\sec\theta+\tan\theta}$$ $$y'=\frac{\sec\theta(\sec\theta+\tan\theta)}{\sec\theta+\tan\theta}=\sec\theta$$
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