Why should the equality of mixed partials be "intuitively obvious"? I am reading Ted Shifrin's excellent book Multivariable Mathematics. It claims that the equality of mixed partials is "an intuitively obvious result, but the proof is quite subtle". However, I guess I must be thinking in the wrong way, because I do not see the intuition behind this result. This is how I think about it:
Let $f:\mathbb{R}^2 \to \mathbb{R}$. I think of $f_x$ as a "field of slopes" in the $x$-direction. If we analyze the movement in the $y$ direction in this field of slopes, we get $f_{xy}$. Now $f_y$ is a "field of slopes" in the $y$-direction. If we analyze movement in the $x$ direction here, we get $f_{yx}$. 
It's unclear to me why movement in the $x$-direction in the "field of $y$-slopes" should be the same as movement in the $y$-direction in the "field of $x$-slopes".
 A: I guess most people develop intuition based on examples, and most examples we pick to examine are $C^2$ functions, where the equality holds. Or, alternatively, you could say that the intuition comes from experience with Taylor's Theorem (which appears in Section 3 of Chapter 5 of my book). The intuition I guess I'm fondest of appears in Chapter 7 (exercise 19 of Section 2), just using a double integral and interchanging the order of integration. (After all, it's natural to think about $\displaystyle{\int\left(\int \frac{\partial^2f}{\partial x\partial y}dy\right)dx}$ and its companion.) I agree that it's not obvious a priori that the $y$ rate of change of $f_x$ should agree with the $x$ rate of change of $f_y$; the $C^2$ condition is subtle, as I said.
A: If you write the difference quotient for a small change $\Delta x$ in $x$ and then the difference quotient for that when you change $y$ by $\Delta y$ the result is the symmetric expression
$$\frac{
f(x + \Delta x, y + \Delta y)
-f(x + \Delta x, y )
-f(   x, y + \Delta y)
+f(x,y)
}
{\Delta x \Delta y} .
$$
