$\begin{align}  & y={{\sin }^{-1}}x \\ & x=\sin y \\ & \text{Differentiate with respect to }x \\ & 1=\cos y\frac{dy}{dx} \\ & \frac{dy}{dx}=\frac{1}{\cos y}=\frac{1}{\sqrt{1-{{\sin }^{2}}y}}=\frac{1}{\sqrt{1-{{(\sin y)}^{2}}}} \\ & \frac{dy}{dx}=\frac{1}{\sqrt{1-{{x}^{2}}}} \\\end{align}$