Answer
$\dfrac {\left( 2\sin x+2x\cos x\right) \left( 3x-1\right) +x\sin x}{\sqrt {3x-1}}=\dfrac {9x\sin x+6x^{2}\cos x-2\sin x-2x\cos x}{\sqrt {3x-1}} $
Work Step by Step
$\dfrac {d}{dx}\left( 2x\left( \sin x\right) \sqrt {3x-1}\right) =\left( \dfrac {d}{dx}2x\sin x\right) \times \sqrt {3x-1}+\left( \dfrac {d}{dx}\sqrt {3x-1}\right) \times 2x\sin x=\left( 2\sin x+2x\cos x\right) \times \sqrt {3x-1}+\dfrac {1}{2}\dfrac {3}{\sqrt {3x-1}}\times 2x\sin x= \dfrac {\left( 2\sin x+2x\cos x\right) \left( 3x-1\right) +3x\sin x}{\sqrt {3x-1}}=\dfrac {9x\sin x+6x^{2}\cos x-2\sin x-2x\cos x}{\sqrt {3x-1}} $
