Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 4: Applications of Derivatives - Section 4.7 - Antiderivatives - Exercises 4.7 - Page 239: 68



Work Step by Step

$\dfrac{d}{dx}[\dfrac{\sin (x^2)}{x}+C] =\dfrac{2x^2 \cos x^2 -\sin x^2}{x^2}$ and $\dfrac{2x^2 \cos x^2 -\sin x^2}{x^2} \ne \dfrac{x \cos x^2 -\sin x^2}{x^2} $ Hence, the given statement is false.
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