If $\mathrm{y}=f(\mathrm{sinx})$ and $f\text{'}(x)={\sin }^{-1} x$ then $\frac{dy}{dx}=$

The given function is $y=f\left(\sin x\right)$ and $f\text{'} (x)={\sin }^{-1} x$

Differentiate both sides with respect to $x$

$\begin{array}{rcl}\frac{dy}{dx}& =& f\text{'}\left(\sin x\right)\cos x\\ & =& {\sin }^{-1}\left(\sin x\right)\cos x\\ & =& x\cos x\end{array}$

Therefore, $\frac{dy}{dx}=\boxed{x\cos x}$