# Partial Derivatives (why do they behave commutatively here)?

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?)

• What is with the $expr$? – smcc Jul 24 '16 at 15:09
• Equality of Mixed Partials is guaranteed when the first partials are continuously differentiable. But that is not a necessary condition. – Mark Viola Jul 24 '16 at 15:10
• nothing I just wanted to spread out the parentheses :) – Conor Cosnett Jul 24 '16 at 15:10
• No, I meant why write $expr=f(x,t)$? – smcc Jul 24 '16 at 15:13

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.)