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: 16

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 $
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