If f is a bijection satisfying f'(x)=sqrt1-f(x)^2, then ≤ft(f^-1)'(x)

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

$⇒f^{-1}of(x)=x⇒f^{-1}(f(x))=x$

so differentiating wrt x

$\frac{df^{-1}(f(x))}{dx}=\frac{1}{\frac{df(x)}{dx}}=\frac{1}{\sqrt{1-\{f(x)\}^2}}$

so $(f^{-1})'(x)=\frac{1}{\sqrt{1-x^2}}$