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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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    @ZhenLin - sorry, I was unaware of the notation2012-04-22

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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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    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!2017-04-23
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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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    Sorry! I had a typo. it's fixed now2012-04-21