Derivative of $F^{-1}(x)$ where $F(x)=\int_{0 }^x\frac{1}{\sqrt{1-t^2}}dt $

Question

Find $(F^{-1})'(x) $ in terms of $F^{-1}(x)$ where $$F(x)=\int_{0 }^x\frac{1}{\sqrt{1-t^2}}dt $$

I know that $$(F^{-1})'(x) = \dfrac{1}{F'(F^{-1}(x))}$$

I calculated using Fundamental Theorem of Calculus, $$F'(x)=\frac{1}{\sqrt{1-x^2}} $$ But I don't know how to proceed

Answer 1

$$F(x)=\int_{0 }^x\frac{1}{\sqrt{1-t^2}}dt $$ By $t=\sin \theta$, $$ F(x) = \int_0^{\sin^{-1}(x)} \frac{\cos \theta d\theta}{\sqrt{1-\sin^2\theta}} = \int_0^{\sin^{-1}(x)} d\theta = \sin^{-1}(x) $$ Because $$(F^{-1})'(x) = \dfrac{1}{F'(F^{-1}(x))} \quad \text{and} \quad F'(x)=\frac{1}{\sqrt{1-x^2}}$$ So,

$$ (F^{-1})'(x) = \sqrt{1-\sin^2 x} $$

Answer 2

Since $F'(x)=\frac1{\sqrt{1-x^2}}$,$$(F^{-1})'(x)=\frac1{F'\bigl(F^{-1}(x)\bigr)}=\sqrt{1-\bigl(F^{-1}(x)\bigr)^2}.$$