Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.7 The Chain Rule - 3.7 Exercises: 70

Answer

$\frac{d^2y}{dx^2}[xcos(x^2)]= -6xsin(x^2)-4x^3cos(x^2)$

Work Step by Step

Chain Rule $\frac{d}{dx}[f(g(x))] = f'(g(x)) \times g'(x)$ $y=xcos(x^2)$ Product Rule and Chain Rule: $\frac{d}{dx}[xcos(x^2)] = cos(x^2) + x(2x(-sin(x^2))) = cos(x^2)-2x^2sin(x^2)$ Product Rule and Chain Rule: $\frac{d^2}{dx^2}[cos(x^2)-2x^2sin(x^2)] = [-2xsin(x^2)]+[-4xsin(x^2)-2x^2(2x)cos(x^2)]=[-2xsin(x^2)]+[-4xsin(x^2)-4x^3cos(x^2)] = -6xsin(x^2)-4x^3cos(x^2)$
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