Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Review - Exercises: 34

Answer

$y'=(\pi\ln10)(\sec^{2}\pi\theta)(10^{\tan\pi\theta})$

Work Step by Step

$y=10^{\tan\pi\theta}$ Start the differentiation process by using the chain rule: $y'=(10^{\tan\pi\theta})(\ln10)(\tan\pi\theta)'=...$ Use the chain rule again to evaluate the derivative indicated and simplify: $...=(10^{\tan\pi\theta})(\ln10)(\sec^{2}\pi\theta)(\pi\theta)'=...$ $...=(\pi\ln10)(\sec^{2}\pi\theta)(10^{\tan\pi\theta})$
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