Let $f(x)=\sin{x}$, $g(x)=f^{-1}(x)=\arcsin{x}$. Then $f'(x)=\cos{x}$,

$$g'(x)=\frac{1}{f'(g(x))}=\frac{1}{\cos(\arcsin{x})}$$

Let $u=\arcsin{x}$, then 

$$x=\sin{u}\implies \cos(\arcsin{x})= \cos{u} = \sqrt{1-x^2}$$

$$g'(x)=\boxed{\frac{d}{dx} \arcsin{x}=\frac{1}{\sqrt{1-x^2}}}$$