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.
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