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I encountered this in a derivation of the 1D wave equation.

Why does the order of application not matter?

$$ expr = f(x,t)$$

$$ \frac{\partial }{\partial t} \left( \frac{\partial }{\partial x } \left(expr \right ) \right) = \frac{\partial }{\partial x} \left( \frac{\partial }{\partial t } \left(expr \right ) \right) $$

why is this true?

(Is it also true for Total Derivatives in the same situation?)

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    $\begingroup$ What is with the $expr$? $\endgroup$ – smcc Jul 24 '16 at 15:09
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    $\begingroup$ Equality of Mixed Partials is guaranteed when the first partials are continuously differentiable. But that is not a necessary condition. $\endgroup$ – Mark Viola Jul 24 '16 at 15:10
  • $\begingroup$ nothing I just wanted to spread out the parentheses :) $\endgroup$ – Conor Cosnett Jul 24 '16 at 15:10
  • $\begingroup$ No, I meant why write $expr=f(x,t)$? $\endgroup$ – smcc Jul 24 '16 at 15:13
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By Schwarz's/Clairaut's/Young's Theorem $$f_{xt}(x,t)=f_{tx}(x,t)$$ or in your notation

$$\frac{\partial^2 f}{\partial t\partial x}=\frac{\partial^2 f}{\partial x\partial t}$$

if $f$ is twice continuously differentiable. (A weaker sufficient condition for symmetry of second-order partial derivatives is that all first-order partial derivatives are differentiable.)

About your question regarding "total derivatives" if you mean the partial derivatives of a composition of functions then yes, if the composition is twice continuously differentiable.

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  • $\begingroup$ Nice! I now know when I can use this (when my function is twice differentiable...). $\endgroup$ – Conor Cosnett Jul 24 '16 at 15:21

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