Answer
$y'=\dfrac{sin \sqrt x}{\sqrt x}+\cos\sqrt x$
Work Step by Step
Given that $y=2\sqrt x \sin\sqrt x $
Apply derivative rules of differentiation:
$f(x)=p'(x)q(x)+p(x)q'(x)$
$y'=\dfrac{d}{dx}[2\sqrt x \sin\sqrt x]$
$=2[\dfrac{d}{dx}(\sqrt x) \sin\sqrt x+(\sqrt x (\cos\sqrt x)]$
$=2[\dfrac{sin \sqrt x}{2\sqrt x}+\dfrac{\cos\sqrt x}{2}]$
$=2(\dfrac{sin \sqrt x}{2\sqrt x})+2(\dfrac{\cos\sqrt x}{2})$
Hence, $y'=\dfrac{sin \sqrt x}{\sqrt x}+\cos\sqrt x$