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: 17

Answer

$y' = 3x^{2}sinx\ cosx+x^{3}cos^{2}x-x^{3}sin^{2}x$ $y' = x^{2}sin^{2}x(3cotx + xcot^{2}x-x)$

Work Step by Step

$y = x^{3}sinx\ cosx$ $y' = (x^{3}sinx)cosx$ $y' = (3x^{2}sinx + x^{3}cosx)cosx - x^{3}sin^{2}x$ $y' = 3x^{2}sinx\ cosx+x^{3}cos^{2}x-x^{3}sin^{2}x$ $y' = x^{2}sin^{2}x(3cotx + xcot^{2}x-x)$
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