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In single variable calculus knowing that a bijective function $f$ has a non-zero derivative at a point $a$ and its inverse is continuous at the point $f(a)$ is enough to conclude (via the chain rule) that: $$(f^{-1})'(f(a))=\frac{1}{f'(a)}$$

In multivariable calculus it seems that to use a similar approach to conclude the analogous result it is not enough to have continuity of the inverse function at $f(a)$, but we also need to know the the inverse is differentiable at $f(a)$. Is this right or is there some weaker condition I can assume for the inverse function to apply the chain rule?
If I need to check that all the partial derivatives of the inverse function exist and are continuous to use the formula for the derivative of the inverse function, then it is much easier to compute directly the Jacobian using the partials.

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  • $\begingroup$ See en.wikipedia.org/wiki/Inverse_function_theorem. In that, the condition that the differential (total derivative) of a continuously differentiable function be non-singular at a point is enough to guarantee both that an inverse exists in a neighborhood of the point and that the inverse is itself differentiable. $\endgroup$ – amd Jan 13 '17 at 21:29

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