Inverse of a bijection f is equal to its derivative
Does there exist a differentiable bijection $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f'(x) = f^{-1}(x)$ ?
No such bijection exists.
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.