pl can some one differentiate y=sin(sin(log 3x))

Given, 
y = sin(sin(log3x))
​Now differentiating both side we get, 

'$$\begin{gathered} \frac{dy}{dx }=\frac{d}{dx }\left(\sin \left(\sin \right(\log 3x\left)\right)\right) \\ \mathrm{We} \mathrm{know} \mathrm{that} \frac{d}{dx }\sin x= \cos x so, \\ \frac{dy}{dx }=\cos \left(\sin \right(\log 3x\left)\right)\times \frac{d}{dx }\sin \left(\log 3x\right) \\ =\cos \left(\sin \right(\log 3x\left)\right)\times \cos \left(\log 3x\right)\times \frac{d}{dx }\log 3x \\ \mathrm{Now} \mathrm{we} \mathrm{know} \mathrm{that} \log \mathrm{x} = \frac{1}{\mathrm{x} } \mathrm{so}, \\ \frac{dy}{dx }=\cos \left(\sin \right(\log 3x\left)\right)\times \cos \left(\log 3x\right)\times \frac{1}{3x}\frac{d}{dx }3x=\frac{\cos \left(\sin \right(\log 3x\left)\right)\times \cos \left(\log 3x\right)}{x}\mathbf{Answer}\mathbf{.} \end{gathered}$$