If f is a bijection satisfying $f'(x)=\sqrt{1-\{f(x)\}^2}$, then $\left(f^{-1}\right)'(x)$

is equal to $\frac{1}{\sqrt{1-x^2}}$ may not exist for every $x \in R$ may not be known explicitly is equal $\sin ^{-1}(f(x))$

is equal to $\frac{1}{\sqrt{1-x^2}}$

$f'(x)=\sqrt{1-\{f(x)\}^2}$ $⇒f^{-1}of(x)=x⇒f^{-1}(f(x))=x$ so differentiating wrt x $\frac{df^{-1}(f(x))}{dx}=\frac{1}{\frac{df(x)}{dx}}=\frac{1}{\sqrt{1-\{f(x)\}^2}}$ so $(f^{-1})'(x)=\frac{1}{\sqrt{1-x^2}}$