Derivative of $\arcsin(x)$

Because the rule is$$(f^{-1})'(x)=\frac1{f'\bigl(f^{-1}(x)\bigr)}$$and therefore\begin{align}\arcsin'(x)&=\frac1{\cos\bigl(\arcsin(x)\bigr)}\\&=\frac1{\sqrt{1-\sin^2\bigl(\arcsin(x)\bigr)}}\\&=\frac1{\sqrt{1-x^2}}.\end{align}

In fact, we have

$ (\arcsin x)' = \dfrac{1}{\sqrt{1-x^2}}$

since,$(f^{-1}(x))' = \dfrac 1{f'(y)}$

with $$sin(y)=x$$ we get $$\cos(y)\frac{dy}{dx}=1$$ so $$\frac{dy}{dx}=\frac{1}{\cos(y)}$$ therefore $$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$$