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: 16

Answer

The Derivative is: $y'=-x^2\sin x$

Work Step by Step

$y=x^2\cos x- 2x\sin x -2\cos x$ Applying Derivative rules: $y'=f'(x)+g'(x)$ and $h'(x)=v'(x)\cdot u(x)+v(x)\cdot u'(x)$ $y'=((2)x^{2-1}(\cos x)+x^2(-\sin x))-2(x^{1-1}(\sin x)+x(\cos x))-2(-\sin x)$ $y'=x\cos x-x^2\sin x-2\sin x-x\cos x+2\sin x$ $y'=-x^2\sin x$
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