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,Assume $f$ a bijective and differentiable function on its domain , I want to find solutions of this functional such that $f\colon\mathbb{R} \to \mathbb{R}$; $f'(f(x))=\exp(f'^{-1}(x))$ such that $f'^{-1}$ is a compositional inverse of $f'$ , I have tried $f(x)=-x$ seems works because we have $f'(f(x))$ is increasing function in the same time $f^{-1}$ exist and would be increasing imply $f'^{-1}$ exist and increasing , if we raise exponential we have increasing functions, but am not sure about this solution ? Is there a clear solution, because its seems that it has a nontrivial solution which it is a formal power series arround $x=0 $ , we may get its coefficients using inverse function theorem ?

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  • $\begingroup$ Should $f'^{-1}$ be the inverse of $f$ or of $f'$? You say two different things. Your attempt seems to suggest $f'^{-1}$ is the inverse of $f$, but the notation suggests otherwise. $\endgroup$ – Batominovski Apr 26 at 19:49
  • $\begingroup$ Compositional inverse of f' , just a wrong typo $\endgroup$ – zeraoulia rafik Apr 26 at 19:50

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