# Inverse of a bijection f is equal to its derivative

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

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Motivated by your question, I asked the following question, in case you might be interested. –  Dejan Govc Jan 15 '13 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.