How to explain Clairaut-Schwartz's Theorem, $f_{xy}=f_{yx}$? I am looking for a non-technical explanation of Clairaut's theorem which states that the mixed derivative of smooth functions are equal.
A geometrical, graphical, or demo that explains the theorem and its implications will be helpful. I am not looking for a proof!
 A: As a first pass, imagine standing on a hill that represents a graph of a function from $\mathbb{R}^2$ to $\mathbb{R}$. Suppose you're facing in the positive $x$ direction, with the positive $y$ direction to your left. There are slopes in both directions: imagine taking a cross section of the hill in either the $x$ or $y$ direction and finding the slope as in single-variable calculus. 
$f_{xy}$ is the amount that the $x$ slope changes as you move in the $y$ direction. $f_{yx}$ is the amount that the $y$ slope changes as you move in the $x$ direction. Playing with this picture in your head for a while should provide some plausibility to the claim that they are the same.
A: I don't think that there is a simple geometric or physical argument that makes this theorem intuitively obvious. In the following I try to explain what kind of information about $f$ the mixed partials do encode. That they are equal then follows from symmetry.
Consider the small square $Q_h:=[-h,h]^2$ and add up the values of $f$ at the vertices of $Q_h$, albeit with alternating signs:
$$\Phi(h):=f(h,h)-f(-h,h)+f(-h,-h)-f(h,-h)\ .$$
Assume that the mixed partial derivatives of $f$ are defined and continuous in a neighborhood of  $(0,0)$. Then we can write
$$\eqalign{\Phi(h)&=\int_{-h}^h f_x(t,h)\>dt-\int_{-h}^h f_x(t,-h)\>dt=\int_{-h}^h\int_{-h}^h f_{xy}(t,s)\>ds\>dt\cr &= f_{xy}(\tau,\sigma)|Q_h|\cr}$$
for some point $(\tau,\sigma)\in Q_h$. Doing the same calculation the other way around we obtain
$$\Phi(h)= f_{yx}(\tau',\sigma')|Q_h|$$
for some point $(\tau',\sigma')\in Q_h$. Letting $h\to0+$ the claim follows.
A: Say you are on a sphere on an initial point $x_0$, and you move along the latitude with constant longitude,
then you move along the longitude with respect to constant latitude (from last previous step).
Now consider doing the converse process from the same initial point $x_0$, 
i.e move along the longitude with constant latitude,
then you move along the latitude with respect to constant longitude (from last previous step).
Are you going to drop on the same (final) spot? if yes then the sphere is smooth!
In simple terms, the theorem states that if the manifold does not have major discontinuities (sufficiently smooth), then the order of the derivatives does not matter, else the discontinutities will throw the computations off, if done if different order, thus in general will not have same result.
Check following illustration for an intuitive view. The wall represents a major discontinuity in the manifold (obstacle), so the paths taken from $A$ to $B$ do not land one same final spot because the discontituity (infinite wall obstacle) throws the computations off (blocked)
In mathematical terms, one can represent this as such:
Moving from $f(x_0,y_0)=f^{00}$ to $f(x_0+h,y_0+h)=f^{hh}$ (in an infinitesimal box $[0,h] \times [0,h]$):


*

*Move first through $x$, then $f^{h0}=f(x_0+h,y_0)=f^{00}+hf_x^{00}+0f_y^{00}$

*Then through $y$, ${}_{x,y}f^{hh}=f(x_0+h,y_0+h)=f^{00}+h(f_x^{00}+f_y^{00}) + h^2f_{yx}^{00}$

*Move first through $y$, then $f^{0h}=f(x_0,y_0+h)=f^{00}+hf_y^{00}+0f_x^{00}$

*Then through $x$, ${}_{y,x}f^{hh}=f(x_0+h,y_0+h)=f^{00}+h(f_x^{00}+f_y^{00}) + h^2f_{xy}^{00}$


The difference between ${}_{y,x}f^{hh}, \; {}_{x,y}f^{hh}$ is $h^2(f_{xy}^{00}-f_{yx}^{00})$, thus if $f_{xy} = f_{yx}$, one will land on same final spot (for example see below).

(in category-theoretic terms, one can say that the "diagram does not commute" :))
note the above diagram is a loose analogy, but one can easily make it into a direct analogy, for example, by adding springs or constraints, which change the amount needed to travel in each direction, given the other direction is already travelled by an amount $h$.
