Answer
False
Work Step by Step
We have: $\dfrac{d}{dx}[\dfrac{\sin (x^2)}{x}+C]$
$\dfrac{d}{dx}[\dfrac{\sin (x^2)}{x}+C] =\dfrac{2x^2 \cos x^2 -\sin x^2}{x^2} \ne \dfrac{x \cos x^2 -\sin x^2}{x^2} $
Hence, the statement is false.
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 273: 88
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