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I know that $\frac {\partial^2 f}{\partial x \partial y} = \frac {\partial^2 f}{\partial y \partial x}$ when $\frac {\partial f}{\partial x}$, $\frac {\partial f}{\partial y}$ exist and are continuous, via the Schwarz's theorem.

I wonder if this logic is valid for higher order derivatives or not, such as:

$\frac {\partial^3 f}{\partial x \partial y \partial z} = \frac {\partial^3 f}{\partial x \partial z \partial y} = \frac {\partial^3 f}{\partial y \partial x \partial y} = \frac {\partial^3 f}{\partial y \partial z \partial y} = \frac {\partial^3 f}{\partial z \partial x \partial y} = \frac {\partial^3 f}{\partial z \partial y \partial x}$

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    $\begingroup$ Hint: Consider $g=\frac{\partial f}{\partial x}$. $\endgroup$
    – Dispersion
    Apr 23, 2019 at 20:47

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Yes it is true, you could see it as taking the $x$ derivative of the expression $f_{zy} = f_{yz}$ for example, or giving the original function as $f_x$

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