Answer
$y' = -x^{2}sinx $
Work Step by Step
$y = x^{2}cosx - 2xsinx -2cosx$
$y' = 2xcosx -x^{2}sinx - 2sinx - 2xcosx +2sinx$
$y' = -x^{2}sinx $
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 3 - Section 3.5 - Derivatives of Trigonometric Functions - Exercises - Page 151: 16
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