Proving $(f^{-1}(x))'\cdot f'(f^{-1}(x))\equiv 1$

Question

I was playing around with an integral formula I came across recently: $$\int g'(x)f(g(x))f'(g(x))dx=\frac12f^2(g(x))+C$$ I decided to try $g(x)=f^{-1}(x)$ to see what happens.

Note I'm assuming inverses and differentials exist for the function

It gives: $$\int (f^{-1}(x))'\cdot x\cdot f'(f^{-1}(x)) dx=\frac12f^2(f^{-1}(x))+C=\frac{x^2}{2}+C$$ But, $$\int x dx =\frac{x^2}{2}+C$$ so this implies that the rest of the integrand is equivalent to $1$, but I can't seem to find a way to show that fact directly, without the use of the integral formula.

Answer 1

You have $$ f(f^{-1}(x))= x $$ for all $x$ where the inverse function is defined. Differentiating this identity with respect to $x$ and using the chain rule gives $$ (f^{-1}(x))'\cdot f'(f^{-1}(x))= 1\, . $$ (This is how the rule for differentiating an inverse function is usually proved.)

Answer 2

$(f^{-1}(x))'=\frac 1 {f'(f^{-1}(x))}$ (assuming that $f$ is differentiable and $f'(f^{-1}(x)) \neq 0$) and this formula can be found in almost any Calculus text.

Proof: Just differentiate $f(f^{-1}(x))=x$ using Chain Rule.

Answer 3

Let $g(x)=f^{-1}(x)$ $$\Rightarrow f(g(x))=f(f^{-1}(x))=x$$ Differentiating both side w.r.t x we get $$f'(g(x)).g'(x)=1$$ $$\Rightarrow f'(f^{-1}(x)).(f^{-1}(x))'=1$$