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In ${\mathbb R}^n$, let $F$ be a smooth one-to-one mapping of $\Omega$ onto some open set \Omega', where $\Omega\subset{\mathbb R}^n$ is open. Set $y=F(x)$. Assume that the Jacobian matrix $J_x=[(\partial y_i/\partial x_j)(x)]$ is nonsingular for $x\in\Omega$. We have $\frac{\partial}{\partial x_j}=\sum\frac{\partial y_i}{\partial x_j}\frac{\partial}{\partial y_i}.$

Here are my questions:

Is there a neat way to calculate $\frac{\partial^2}{\partial x_j\partial x_k}?$ After several steps trial, I am completely confused. More generally, what is $\partial_x^{\alpha}$ in terms of the $y$ coordinate system? Here $\partial_x^{\alpha}:=\frac{\partial^{|\alpha|}}{\partial x_1^{\alpha_1}\cdots\partial x_n^{\alpha_n}}.$

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    Lik$e$ a$n$on says, th$e$ "difficulty" is already present in the one-dimensional case. You should think about that before you deal with change of coordinates in higher dimensions.2011-08-12

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It's a straightforward computation, you just have to do it: $\frac{\partial^2f}{\partial x_j\partial x_k}=\frac{\partial}{\partial x_j}\left(\sum_i\frac{\partial y_i}{\partial x_k}\frac{\partial f}{\partial y_i}\right)=\sum_i\frac{\partial^2 y_i}{\partial x_j\partial x_k}\frac{\partial f}{\partial y_i}+\sum_i\frac{\partial y_i}{\partial x_k}\frac{\partial}{\partial x_j}\left(\frac{\partial f}{\partial y_i}\right)$

Now all that is left is to also replace $\frac{\partial}{\partial x_j}$ so you get

$\frac{\partial^2}{\partial x_j\partial x_k}=\sum_i\frac{\partial^2 y_i}{\partial x_j\partial x_k}\frac{\partial}{\partial y_i}+\sum_{h,i}\frac{\partial y_i}{\partial x_k}\frac{\partial y_h}{\partial x_j}\frac{\partial^2}{\partial y_h\partial y_i}$ Notice that $\frac{\partial}{\partial x_j}$ is a derivation, so it can be defined by a (coordinate independent) vector field.