# Why does $\frac{\partial^2f}{\partial x \partial y} = \frac{\partial^2f}{\partial y \partial x}$

I'm studying multivariable calculus and I came across the following property of partial derivatives:

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

where $f = f(x,y)$. Somehow this seems intuitive, but if I would need to prove this to a 10-year-old what should I do? So I'm looking for the proof :) Is this true in the general case? For example if $f = f(x_1, ..., x_n)$ and you would take the partials in random order. Would the rule still hold?

Thank you for any help :)

R.T.P

$\frac { \partial } {\partial x} [\frac { \partial } {\partial y} f ]= \frac { \partial } {\partial y} [\frac { \partial } {\partial x} f ]$ .... (1)

Let

$L(x,y) = \frac { \partial } {\partial y} f = \lim_{h\to 0} \frac{f(x,y+h) - f(x,y)}{h}$

$M(x,y) = \frac { \partial } {\partial x} f = \lim_{h\to 0} \frac{f(x+h,y) - f(x,y)}{h}$

$\frac{\partial}{\partial x} L = \lim_{h\to 0} \frac{L(x+h,y) - L(x,y)}{h}$

$\frac{\partial}{\partial y} M = \lim_{h\to 0} \frac{M(x,y+h) - M(x,y)}{h}$

$\frac{\partial}{\partial x} L = \lim_{h_1\to 0} \lim_{h_2 \to 0} \frac{(\frac{f(x+h_1,y+h_2) - f(x+h_1,y)}{h_2}) - (\frac{f(x,y+h_2) - f(x,y)}{h_2})}{h_1}$ ... (2)

$\frac{\partial}{\partial y} M = \lim_{h_1\to 0} \lim_{h_2\to 0} \frac{(\frac{f(x+h_2, y+h_1) - f(x,y+h_1)}{h_2}) - (\frac{f(x+h_2, y) - f(x,y)}{h_2})}{h_1}$ ... (3)

(2) and (3) are equivalent which proves (1)

See here

• This is very incomplete. The equality is not true in general, and I can't see what kinds of assumptions you put on $f$ and where you use them. – mrf Aug 15 '13 at 10:18
• I think you're right, now that you mention it I think I need to at least use continuity in there somewhere – sav Aug 15 '13 at 10:26
• You need $f$ to be $C^2$, I believe. – Javier Aug 15 '13 at 10:45