# Existence of a function whose derivative of inverse equals the inverse of the derivative

I've been thinking about the calculation of inverse functions through Taylor series expansions.

My hypothesis was that if we had: $$\ f(x) =\sum_{n=0}^\infty \frac{(x-x_0)}{n!}f^{n}(x_0),$$ then $$\ f^{-1}(x) =\sum_{n=0}^\infty \frac{(x-x_0)}{n!}({f^{-1}})^{(n)}(x_0).$$ Here it'd be possible to calculate the derivatives $$\ ({f^{-1}})^{(n)}(x_0)$$ through the inverse function derivation theorem but for this to work we should have the derivative of the inverse of$$\ f$$ equal the inverse of the derivative of $$\ f$$, i.e. $$\ (f^{-1})' = (f')^{-1} \quad (1).$$ In general this does not hold; take for example $$\ f(x) = x^2$$.

To formulate my question, assume we have a differentiable bijective map $$\ f:A\rightarrow B$$ with bijective derivative. Its inverse $$\ f^{-1}$$ is also a differentiable bijection. Assume the derivative of the inverse is also bijective.

To try and find a function for which$$\ (1)$$ holds, I was able to deduce (from$$\ (1)$$) that if such a function exists, we must have $$\ f(x) = (f' \circ f')(x) \quad (2).$$

Clearly this does not hold for any polynomial nor (I'm pretty sure) for any other elementary functions. So, does there exist a function for which this condition holds?

Deriving (2): $$\ (f')^{-1}(x) = (f^{-1})'(x) = \frac{1}{(f' \circ f^{-1})(x)} \iff (f' \circ f^{-1})(x) = \frac{1}{(f')^{-1}(x)}$$. Then mapping by $$\ f$$ from the right gives $$\ f'(x)=\frac{1}{(f')^{-1}(f(x))}$$ and mapping by $$\ \frac{1}{f'}$$ from the left $$\ \frac{1}{f(x)} = (\frac{1}{f'} \circ f')(x) \iff f(x)= \frac{1}{(\frac{1}{f'} \circ f')(x)}=\frac{1}{\frac{1}{(f' \circ f')(x)}}=(f' \circ f')(x)$$

• How did you derive the fact $f(x)=(f' \circ f')(x)?$ – JasonM Jun 17 '16 at 8:43
• @JasonM I edited the post now. – JTervonen Jun 17 '16 at 9:26
• Actually, I'm feeling a bit uneasy about the step where you mapped by $\frac{1}{f'}$ from the left. Are we sure $(\frac{1}{f'}) \circ \frac{1}{((f')^{-1})\circ f}=\frac{1}{f}$? – JasonM Jun 17 '16 at 10:01
• Ah yes, I've mixed multiplication and mapping there. Actual mapping or multiplication don't seem to yield anything useful either. I'll try and work on this a bit more. – JTervonen Jun 17 '16 at 10:36

## 1 Answer

Let $f(x)$ be such invertible function s.t. $(f^{-1})'={f'(x)}^{-1}$ which gives a non-linear differential equation $f'^2 +f^2=0$. Clearly, $f=0$ is the only solution to it, which contradicts that $f(x)$ is invertible.

• Could you explain to me how you got the expression $f'^2 + f^2 = 0$? – Fabian Schn. Aug 8 '19 at 11:01
• $LHS=(f^{-1})'=(\frac{1}{f})'=-\frac{f'(x)}{f^2(x)}$ and $RHS=\frac{1}{f'(x)}$. – Nitin Uniyal Aug 9 '19 at 0:44
• Why is $f^{-1}=\frac{1}{f}$? Wouldn't that imply that $f(x)=\frac{1}{x}$? Also, I don't see how $(f')^{-1}=\frac{1}{f'}$. – Void Apr 10 '20 at 3:03