# Geometric or intuitive proof of the symmetry of second partial derivatives

What was given in my calc book is a "consider the function" proof. That is, the author gives a function out of the blue and would deduce all the nice properties from it. I'd prefer a proof which is motivated (perhaps, intuitive) - you see how the proof is crafted in the mind of the person. So my question is a geometric or intuitive proof of

$$\frac{\partial ^2 f}{\partial x \, \partial y} = \frac{\partial^2 f}{\partial y \, \partial x}$$

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A (perversely) related question. – J. M. Jul 14 '12 at 15:56
This link has a nice proof, and probably is the "consider this function" proof you have in your book, but the remark does a good job of saying why we consider such a function, and brings out the heart of the problem. Usually the intuitive part is that they should be equal, and the bit that denies us is that they can be unequal. Equal of mixed partials comes down to the ability to interchange a pair of limits and when you see this problem in this light it is easier to construct a counterexample. – Ragib Zaman Jul 14 '12 at 16:06

One intuitive explanation for this is the following simple geometrical fact: if you first shift the $xy$-plane to the right and then shift it up, you end up with the same result as if you had first shifted everything up and then to the right.

Using some fancy mathematical language: the equation $\frac{\partial}{\partial x}\frac{\partial}{\partial y}=\frac{\partial}{\partial y}\frac{\partial}{\partial x}$ says that the standard vector field in $x$ direction commutes with the standard vector field in $y$ direction (their Lie bracket is zero). One can prove that this is the same as saying that their flows commute which is the statement I gave above.

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For monomial functions of the form $f(x,y) = x^a y^b$ (where $a$ and $b$ are nonnegative integers), it is easy to check this result directly. Since the result is linear in $f$, it also holds for polynomial $f$.

Any smooth-enough function $f$ (for example, if $f$ has continuous second partials in the neighborhood of a point $P$) is "approximately" a polynomial in a neighborhood of $P$, so the result will also hold for such $f$ at the point $P$.

The "consider the function" proof you mention from your textbook is probably just a formalization of this idea. If you write out that proof, perhaps I (or others) can help you see that.

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It boils down to the algebraic fact that it is all the same in that order to take coordinate differences: $$\Delta_x(h) \Delta_y(h)f(x,y)= \Delta_x(h)(f(x,y+h)-f(x,y))=$$ $$f(x+h,y+h)-f(x+h,y)-f(x,y+h)+f(x,y)=$$ $$\Delta_y(h)(f(x+h,y)-f(x,y))=\Delta_y(h)\Delta_x(h) f(x,y).$$ Then a calculus reasoning follows that under certain conditions $$\dfrac{\partial ^2 f}{\partial x \partial y} = \lim_{h\to0}\frac{\Delta_y(h)\Delta_x(h) f(x,y)}{h^2}$$ and $$\dfrac{\partial^2 f}{\partial y \partial x} = \lim_{h\to0}\frac{\Delta_x(h)\Delta_y(h) f(x,y)}{h^2}.$$

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