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: 71

Answer

$\frac{d}{dx}sin(x^2) = 2cos(x^2)-4x^2sin(x^2)$

Work Step by Step

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