Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

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

Answer

The Derivative is: $y'=-\tan^2 x+4\tan x\sec^2 x-2x\tan x\sec^2 x$

Work Step by Step

$y=(2-x)\tan^2 x$ $y=(2-x)\tan x\tan x$ Applying Derivative rules: $y'=f'(x)\cdot g(x)+f(x)\cdot g'(x)$ $y'=(0-x^{1-1})\tan x\tan x+(2-x)((sec^2 x)\tan x+\tan x(\sec^2x))$ $y'=(-1)\tan^2 x+(2-x)(2\tan x\sec^2 x)$ $y'=-\tan^2 x+4\tan x\sec^2 x-2x\tan x\sec^2 x$
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