Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Review - Exercises: 33

Answer

$y'=5\sec5x$

Work Step by Step

$y=\ln|\sec5x+\tan5x|$ Start the differentiation process by using the chain rule: $y'=\dfrac{(|\sec5x+\tan5x|)'}{|\sec5x+\tan5x|}=...$ Apply the chain rule again to find the derivative indicated in the numerator: $...=\dfrac{\Big(\dfrac{\sec5x+\tan5x}{|\sec5x+\tan5x|}\Big)(\sec5x+\tan5x)'}{|\sec5x+\tan5x|}=...$ $...=\dfrac{(\sec5x+\tan5x)(\sec5x+\tan5x)'}{|\sec5x+\tan5x|^{2}}=...$ Use the chain rule one last time to evaluate the derivative indicated in the numerator and simplify: $...=\dfrac{(\sec5x+\tan5x)[(\sec5x\tan5x)(5x)'+(5x)'\sec^{2}5x]}{|\sec5x+\tan5x|^{2}}=...$ $...=\dfrac{(\sec5x+\tan5x)[5(\sec5x\tan5x)+5\sec^{2}5x]}{|\sec5x+\tan5x|^{2}}=...$ $...=\dfrac{5(\sec5x+\tan5x)(\sec5x\tan5x+\sec^{2}5x)}{|\sec5x+\tan5x|^{2}}=...$ Take out common factor $\sec5x$ from the numerator: $...=\dfrac{5\sec5x(\sec5x+\tan5x)(\sec5x+\tan5x)}{|\sec5x+\tan5x|^{2}}=...$ $...=\dfrac{5\sec5x(\sec5x+\tan5x)^{2}}{|\sec5x+\tan5x|^{2}}$ =$5\sec5x$
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