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: 9

Answer

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

Work Step by Step

$y=\dfrac{x}{2-\tan x}$ Differentiate using the quotient rule: $y'=\dfrac{(2-\tan x)(x)'-(x)(2-\tan x)'}{(2-\tan x)^{2}}=...$ $...=\dfrac{(2-\tan x)(1)-(x)(-\sec^{2}x)}{(2-\tan x)^{2}}=...$ $...=\dfrac{2-\tan x+x\sec^{2}x}{(2-\tan x)^{2}}$
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