Differentiate $\sin ^{-1}\left\{\frac{2^{x+1} \cdot 3^{x}}{1+(36)^{x}}\right\}$ with respect to $x$.
$y=\sin ^{-1}\left\{\frac{2^{x+1} \cdot 3^{x}}{1+(36)^{x}}\right\}$
$y=\sin ^{-1}\left\{\frac{2 \times 2^{x} \times 3^{x}}{1+\left(6^{2}\right)^{x}}\right\}$
$y=\sin ^{-1}\left\{\frac{2 \times 6^{x}}{1+\left(6^{x}\right)^{2}}\right\}$
Put $6^{x}=\tan \theta$
$y=\sin ^{-1}\left\{\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right\}$
Using $\sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$
Now,
$y=\sin ^{-1}(\sin 2 \theta)$
$y=2 \theta$
$y=2 \tan ^{-1}\left(6^{x}\right)$
Differentiating w.r.t $x$, we get
$\frac{d y}{d x}=\frac{d}{d x}\left(2 \tan ^{-1} 6^{x}\right)$
$\frac{d y}{d x}=2 \times \frac{6^{x} \log 6}{1+\left(6^{x}\right)^{2}}$
$\frac{d y}{d x}=\frac{2 \times 6^{x} \log 6}{1+6^{2 x}}$