Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Review - Exercises: 4

Answer

$y'=\dfrac{\sec^{2}x+\cos x\sec^{2}x+\tan x\sin x}{(1+\cos x)^{2}}$

Work Step by Step

$y=\dfrac{\tan x}{1+\cos x}$ Start the differentiation process by using the quotient rule: $y'=\dfrac{(1+\cos x)(\tan x)'-(\tan x)(1+\cos x)'}{(1+\cos x)^{2}}=...$ Evaluate the derivatives indicated in the numerator: $...=\dfrac{(1+\cos x)(\sec^{2}x)-(\tan x)(-\sin x)}{(1+\cos x)^{2}}=...$ $...=\dfrac{\sec^{2}x+\cos x\sec^{2}x+\tan x\sin x}{(1+\cos x)^{2}}$
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