Let $f$ be continuously differentiable $m$ times at $x \in \mathbb{R}$. Does it then follow that $f^{-1}$ (assuming it exists) is also continuously differentiable $m$ times at $f(x)$? This is true for $m=1$ but does it hold for any $m$? Thank you!
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For all the counterexamples, an answer providing a sufficient condition would be nice! – Sarvesh Ravichandran Iyer Nov 16 '20 at 09:56
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The answer is “yes” if $f'(x) \ne 0$: On derivatives of the inverse of a real function. – Martin R Nov 16 '20 at 09:57
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Teresa Lisbon: Go ahead, provide one. The counterexamples are important, as they point out where the mistake is. The author of the post (or you, if you feel like it) can then fix that mistake, and possibly find a positive result. – A. Pongrácz Nov 16 '20 at 09:58
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No, it is not true for $m=1$. $f(x)=x^3$ is infinitely many times differentiable at 0, and its inverse function $\sqrt[3] x$ is not differentiable at 0. There can be a problem with zero derivatives.
A. Pongrácz
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