Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.3 - Derivatives of Trigonometric Functions - 3.3 Exercises - Page 196: 5

Answer

$y'=\sec^{3}\theta+\sec\theta\tan^{2}\theta$

Work Step by Step

$y=\sec\theta\tan\theta$ Differentiate using the product rule: $y'=(\sec\theta)(\tan\theta)'+(\tan\theta)(\sec\theta)'=...$ $...=(\sec\theta)(\sec^{2}\theta)+(\tan\theta)(\sec\theta\tan\theta)=...$ $...=\sec^{3}\theta+\sec\theta\tan^{2}\theta$
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