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Suppose I have $G: \mathbb{R}^n \rightarrow \mathbb{R}^n$ and it is invertible everywhere and the inverse is continuously differentiable sufficiently many times.

For the first partial there is the formula using the Jacobian matrix. I am wondering how can I compute derivatives like $\partial^2/{\partial x_1 \partial x_2}$ or $\partial^3/{\partial^2 x_1^2 \partial x_2}$? Any comments would be appreciated! Thank you.

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  • $\begingroup$ I would recommend differentiating implicitly, but you're still going to need to be careful with higher-order derivatives as symmetric multilinear maps. $\endgroup$ – Ted Shifrin Sep 30 '19 at 23:13
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Hint (non rigorous) : For $n=1$ you can get the second derivative of the inverse as follow :

If y=G(x), then \begin{align*} G(G^{-1}(y))=y \end{align*} Differentiating both sides gives \begin{align*} &\frac{d}{dy}G^{-1}(y) \frac{d}{dx}G(x)=1\\ \Leftrightarrow~&\frac{d}{dy}G^{-1}(y) =\frac{1}{\frac{d}{dx}G(x)} \end{align*} Differentiating both sides again (and keeping in mind that $x=G^{-1}(y)$) you get \begin{align*} \frac{d^2}{dy^2}G^{-1}(y) &= \frac{\frac{d}{dy} G^{-1}(y)}{\frac{d^2}{dx^2} G(x)}\\ &= \frac{1}{\left(\frac{d}{dx} G(x) \right)\cdot\left(\frac{d^2}{dx^2} G(x) \right)} \end{align*} Where the first line is chain rule and the last line is just replacing The previously obtained derivative of $G^{-1}$.

I believe you can continue this reasoning to get other derivatives. In order to get a $n\times n$ argument, you would need to use tensors and multi-linear algebra which I don't know enough to develop here.

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