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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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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. –  Neal Apr 21 '12 at 15:21
    
Sorry, had a typo. it's fixed now –  Belgi Apr 21 '12 at 15:22
    
*give me a sec to fix the tex –  Belgi Apr 21 '12 at 15:23
    
@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 ? –  Belgi Apr 21 '12 at 15:27
    
Nope! Now it's true! (See my revised answer.) –  Neal Apr 21 '12 at 15:28

2 Answers 2

up vote 3 down vote accepted

(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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Sorry, had a typo. it's fixed now –  Belgi Apr 21 '12 at 15:21
    
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... –  Belgi Apr 21 '12 at 15:30
    
They have used the notation "$\partial_{xy}$" $= \partial_x\partial_y$. –  Neal Apr 21 '12 at 15:38

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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Sorry! I had a typo. it's fixed now –  Belgi Apr 21 '12 at 15:28

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