Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - Review - Exercises - Page 197: 25

Answer

$$y'=\frac{2 \sec 2 \theta(\tan 2 \theta-1)}{(1+\tan 2 \theta)^{2}}$$

Work Step by Step

Given $$y=\frac{\sec2\theta }{1+\tan 2\theta }$$ Then \begin{align*} y^{\prime} &=\frac{(1+\tan 2 \theta)(\sec 2 \theta \tan 2 \theta \cdot 2)-(\sec 2 \theta)\left(\sec ^{2} 2 \theta \cdot 2\right)}{(1+\tan 2 \theta)^{2}}\\ &=\frac{2 \sec 2 \theta\left[(1+\tan 2 \theta) \tan 2 \theta-\sec ^{2} 2 \theta\right]}{(1+\tan 2 \theta)^{2}} \\ &=\frac{2 \sec 2 \theta\left(\tan 2 \theta+\tan ^{2} 2 \theta-\sec ^{2} 2 \theta\right)}{(1+\tan 2 \theta)^{2}} \ \text{Using}\quad\left[1+\tan ^{2} x=\sec ^{2} x\right]\\ &=\frac{2 \sec 2 \theta(\tan 2 \theta-1)}{(1+\tan 2 \theta)^{2}} \end{align*}
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