Differentiating this inverse trigonometric function $$\sin^{-1}\left( \frac{2^{x+1}\cdot3^x}{1+36^{x}} \right)$$
Had this question for todays test but still cannot find out how to proceed.

Let
\begin{align}
y&=\sin ^{-1}\left(\frac{2^{x+1}3^x}{1+36^x}\right)\\
 &=\sin ^{-1}\left(\frac{2^x\cdot2\cdot3^x}{1+36^x}\right)\\
 &=\sin ^{-1}\left(\frac{2\cdot6^x}{1+6^{2x}}\right)\\
&=2\tan ^{-1}6^x
\end{align}
Thus
\begin{align}
\frac{dy}{dx}&=\frac{2}{1+6^{2x}}\cdot6^x\log 6\\
\implies \frac{dy}{dx}&=\frac{2\cdot6^x\log 6}{1+36^x}
\end{align}
