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

Question

Math StackExchange, long time reader, first time writer. I have a question on inverse differentiation. The question is: Suppose $f'(x) = \sqrt{1-f(x)^2}$, then $(f^{-1})' (x) = ?$

I had a similar question: Let f be a differentiable one-to-one function such that $f'(x) = 1+[f(x)]^2$ and $f(1)= 3$ then $(f^-1)'(3) = ?$. The answer I got was $1/10$ (this question was easier as all one has to do is take $1/f'$ ($f^{-1}(x) = 1/f'(3) = 1/10$).

I cannot apply the same logic in the former question as we are dealing with a variable, but I am inclined to believe the answer is $2x/\sqrt{1-x^2}$?

Thank you.

Answer 1

$f^{-1}\circ f(x)=x$ implies $(f^{-1})'(f(x))f'(x)=1$ thus

${f^{-1}}'(f(x))=1/f'(x) = 1/\sqrt{1-f^2(x)}$ , thus ${f^{-1}}'(x)=1/\sqrt{1-x^2}$.

Answer 2

The function $f$ maps a certain $x$-interval $I$ strictly increasingly onto some $y$-interval $J$, and $g:=f^{-1}$ maps $J$ back onto $I$. According to the formula for the derivative of the inverse function one has $$g'(y)={1\over f'\bigl(g(y)\bigr)}={1\over\sqrt{1-f\bigl(g(y)\bigr)^2}}={1\over\sqrt{1-y^2}}\qquad(y\in J)\ .$$ Note that $I:=\bigl]-{\pi\over2},{\pi\over2}\bigr[\ $, $J:=]{-1},1[\ $, and $f:=\sin\restriction I$ form an instance of this problem.