Answer
False
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.
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
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
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
