Question

If f is one-one and satisfies $f^{\prime }\left(x\right)=\sqrt{1-{\left(f\left(x\right)\right)}^2}$, then ${\left(f^{-1}\right)}^{\prime }\left(x\right)$

• A. is equal to $\frac{1}{\sqrt{1-x^2}}$
• B. may not exist for every $x\in (0,1)$
• C. may not be known explicitly
• D. is equal to ${\cos }^{-1}\left(f\left(x\right)\right)$

Answer 1

Answer: (A)

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

$f^{\prime }\left(x\right)=\sqrt{1-{\left(f\left(x\right)\right)}^2}$
let $f\left(x\right)=y$
$\Rightarrow \frac{dy}{dx}=\sqrt{1-y^2}$
$\Rightarrow \frac{1}{\sqrt{1-y^2}}dy=dx$
integrating on both sides gives
${\sin }^{-1}y=x+c$
$\Rightarrow f^{-1}\left(y\right)={\sin }^{-1}y-c$
$\therefore f^{-1}\left(x\right)={\sin }^{-1}x-c$
differentiating w.r.t x on both sides
${\left(f^{-1}\right)}^{\prime }\left(x\right)=\frac{1}{\sqrt{1-x^2}}$
Hence, option A.