Given,

$y={\sin }^{-1}\left[\frac{5x+12\sqrt{1-x^2}}{13}\right]$

On putting $x=\sin \theta ,5=r\cos \alpha$ and $12=r\sin \alpha ,$ we get
$y={\sin }^{-1}\left(\frac{r\cos \alpha \sin \theta +r\sin \alpha \cos \theta }{13}\right)$
$={\sin }^{-1}\left(\frac{13\sin \left(\theta +\alpha \right)}{13}\right) ;\left[\because r=\sqrt{25+144}=13\right]$
$=\theta +\alpha$

$={\sin }^{-1}x+{\tan }^{-1}\left(\frac{12}{5}\right) ;\left[\because \tan \alpha =\frac{12}{5}\right]$

Differentiating both sides w.r.t. $x$, we get

$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$