University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.5 - Derivatives of Trigonometric Functions - Exercises: 18

Answer

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

Work Step by Step

$y = (2-x)tan^{2}x$ $y = (2-x)tanx\ tanx$ $y' = (-tanx +(2-x)sec^{2}x)tanx + (2-x)tanx\ sec^{2}x$ $y' = -tan^{2}x + 2sec^{2}x\ tanx- xsec^{2}x\ tanx + 2tanx\ sec^{2}x - xtanx\ sec^{2}x$ $y' = -tan^{2}x +4sec^{2}x\ tanx-2xsec^{2}x\ tanx$ $y' = 2sec^{2}x\ tanx(2-x) - tan^{2}x$
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