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Consider a function: $$f:\mathbb R ^2\to\mathbb R $$ when does $$\dfrac {\partial f(x,t)}{\partial t\partial x}=\dfrac {\partial f(x,t)}{\partial x\partial t}$$

Thinking about it in terms of the limit definition of derivative, and thinking about it as taking slices of surfaces and measuring the slope on the edge, seems to be giving me the feeling that answer is always.

However I have some memories of there being requirements like piecewise continuous, and smooth.

What is the full set of conditions?

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  • $\begingroup$ It's true where the partial derivatives are all continuous. (But not an iff statement as far as I'm aware). $\endgroup$
    – ah11950
    Mar 25, 2014 at 1:44
  • $\begingroup$ ah11950: That sounds like an answer (not a comment). Why don't you post it as an answer? $\endgroup$
    – Frames Catherine White
    Mar 25, 2014 at 1:48
  • $\begingroup$ The point is that the discrete version of your formula, that is $$\Delta_t\Delta_x f = \Delta_x \Delta_t f, $$ where $\Delta_t f(x, t)=f(x, t +h)$ for a fixed increment $h$, is always true. However, to infer from this that partial derivatives commute you need a limiting process and so you need at least some continuity. $\endgroup$
    – Giuseppe Negro
    Mar 25, 2014 at 2:10
  • $\begingroup$ See also this related question $\endgroup$
    – Giuseppe Negro
    Mar 25, 2014 at 2:13

1 Answer 1

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It's true where the mixed partial derivatives are all continuous. (But not an iff statement as far as I'm aware).

An example of a function that fails to satisfy equality of mixed partial derivatives is

$$f(x,y) = \begin{cases} \frac{xy(x^2-y^2)}{x^2+y^2} \; &\text{if}\; (x,y) \neq 0\\ 0 &\text{if}\; (x,y) = 0\end{cases}$$

At the origin we have $f_{xy}(0,0) = 1 \neq -1 = f_{yx}(0,0)$

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  • $\begingroup$ How did you calculate $f_{xy}, f_{yx}$ so easily? Is there some special technique to this? $\endgroup$
    – PythonSage
    Mar 31, 2020 at 23:56

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