# How do I compute higher derivatives of inverse of multivariable functions?

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.

• I would recommend differentiating implicitly, but you're still going to need to be careful with higher-order derivatives as symmetric multilinear maps. – Ted Shifrin Sep 30 '19 at 23:13

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.