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Let $f:\mathbb{R}^2\to\mathbb{R}$ be a smooth function (i.e. have derivatives of all orders).

Is the relation $$\frac{\partial}{\partial x}\frac{\partial}{\partial y}f = \frac{\partial}{\partial y}\frac{\partial}{\partial x}f$$ true?

I studied calculus a while ago and I don't remember if this is true, can someonem confirm ?

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    $\begingroup$ By the way, "commuting" usually refers to changing the order of something. In this context, your use of the word makes it sound like your question is asking whether $\partial_x\partial_yf = \partial_y\partial_xf$, which is certainly true if $f$ is smooth. $\endgroup$
    – Neal
    Apr 21, 2012 at 15:21
  • $\begingroup$ Sorry, had a typo. it's fixed now $\endgroup$
    – Belgi
    Apr 21, 2012 at 15:22
  • $\begingroup$ *give me a sec to fix the tex $\endgroup$
    – Belgi
    Apr 21, 2012 at 15:23
  • $\begingroup$ @Neal I wrote somthing else by mistake, I fixed the question. the examples that were given are not counter examples to what I ment...is it still false ? $\endgroup$
    – Belgi
    Apr 21, 2012 at 15:27
  • $\begingroup$ Nope! Now it's true! (See my revised answer.) $\endgroup$
    – Neal
    Apr 21, 2012 at 15:28

2 Answers 2

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(answer to first question: Certainly not. Try $f(x,y) = x$. Then $\partial_xf = 1$ but $\partial_yf = 0$.)

In your revised question, yes. In fact, we need only require that the second derivatives of $f$ be continuous, that is, we need only require $f$ be of class $C^2$. This is Clairaut's theorem.

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  • $\begingroup$ Sorry, had a typo. it's fixed now $\endgroup$
    – Belgi
    Apr 21, 2012 at 15:21
  • $\begingroup$ I looked at the link, are you sure this implies that partial derivative commute ? From what I see it sais that ∂xy=∂yx and not ∂x∂y=∂y∂x... $\endgroup$
    – Belgi
    Apr 21, 2012 at 15:30
  • $\begingroup$ They have used the notation "$\partial_{xy}$" $= \partial_x\partial_y$. $\endgroup$
    – Neal
    Apr 21, 2012 at 15:38
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    $\begingroup$ Although not as widely known, the function doesn't have to be twice continuously differentiable (meaning that the first partial derivatives are continuously differentiable, meaning that the second partial derivatives are continuous). It's enough if the function is twice differentiable (meaning that the first partial derivatives are differentiable). Of course, it's important here that ‘differentiable’ means something stronger than that the partial derivatives exist! $\endgroup$
    – Toby Bartels
    Apr 23, 2017 at 4:05
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Certainly not; consider $f(x,y) = x+2y$.

You may be thinking of the theorem that, for smooth functions, $$\frac{\partial}{\partial x}\frac{\partial}{\partial y} f = \frac{\partial}{\partial y}\frac{\partial}{\partial x} f.$$

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  • $\begingroup$ Sorry! I had a typo. it's fixed now $\endgroup$
    – Belgi
    Apr 21, 2012 at 15:28

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