University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Practice Exercises - Page 202: 46



Work Step by Step

$y=\ln{(\sec^2\theta})$ On differentiating both sides: $\frac{dy}{d\theta}=\frac{d(\ln{(\sec^2\theta}))}{d\theta}$ $\frac{dy}{d\theta}=\frac{1}{\sec^2\theta}\frac{d({(\sec^2\theta}))} {d\theta}$ $\frac{dy}{d\theta}=\frac{{(2\sec\theta\sec\theta\tan\theta})}{\sec^2\theta}={2\tan\theta}$ $\frac{dy}{d\theta}=2\tan\theta$
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