# Inverse of a bijection f is equal to its derivative

Does there exist a differentiable bijection $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f'(x) = f^{-1}(x)$ ?

• Motivated by your question, I asked the following question, in case you might be interested. Commented Jan 15, 2013 at 20:23

Let $f\colon\mathbb{R}\to\mathbb{R}$ be a differentiable bijection. Then $f$ is increasing or decreasing. If $f$ is increasing, then $f'(x)\ge0$ for all $x\in\mathbb{R}$, while if $f$ is decreasing then $f'(x)\le0$ for all $x\in\mathbb{R}$. If $f'=f^{-1}$, then $f^{-1}(x)\ge0$ or $f^{-1}(x)\le0$ for all $x\in\mathbb{R}$, contradicting the fact that $f$, and hence $f^ {-1}$, is a bijection.
• Possible counterpoint? What about $f(x)=x^{\phi}*\phi^{1-\phi}$ where $\phi$ is the Golden Ratio. (I stole this solution from mathoverflow.net/questions/34052/function-satisfying-f-1-f/…) Commented Apr 22, 2014 at 4:58
• $f$ is a bijection satisfying $f'=f^{-1}$ on $(0,\infty)$. When considered as a bijection fron $\mathbb{R}$ to itself, it has to be defined as $-f(|x|)$ for $x<0$, and then $f'(x)>0$ for all $x$. Commented Apr 23, 2014 at 9:42