geometric meaning of cross partial derivative The geometric meaning of first order derivative is the slope of a curve(or a slice of the surface for two argument functions); While the second order derivative means the "concavity/convexity" of the curve. 
Now I'm wondering whether there's an analog interpretation for cross partial derivatives(maybe just for two argument functions). 
Thank you!
 A: I'm interested in seeing better answers than mine, but here's a thought:
Consider the graph of $f(x,y)$ as a surface in $\mathbb{R}^3$. Imagine standing on the surface over a particular point $(x_0,y_0)$, and face the positive $y$ direction. Consider the $x$-slope at your position; it determines your left-right tilt as you stand there on the surface. Now walk forward a bit and observe the rate of change of the $x$-slope; i.e. the left-right tilt. More precisely, walk along the path lying over $t\mapsto(x_0,y_0+t)$ and observe the rate of change of $t\mapsto f_x(x_0,y_0+t)$. That's the cross partial derivative at $(x_0,y_0)$; it's the rate at which $x$-slices "twist" as you travel in the $y$ direction.
I think a proper answer to your question should fold some instrinsic geometric features of the surface into the discussion, and I'm not sure how to do that.
A: for a function $f(x,y)$ you can think of $f_{xy}$ as the slope wrt $y$ of the slope wrt $x$, or alternatively, the slope wrt $x$ of the slope wrt $y$. It is not intuitive they should be the same, but they turn out to be...
A: Essentially there are two types of points on a surface.
Synclastic.
Concave or convex surface along principal directions of same sign. Gauss curvature  $ K = k_1 k_2  > 0 $ . Concavity at intermediate $ \psi$ meaningful, retains convexity or concavity.
Anticlastic.
Concavity or convexity of  this surface is meaningless. Normal curvature directions have opposite sign. Gauss curvature  $ K= k_1 k_2  < 0 $ .
The concept of normal curvature by Euler and geodesic torsion $\sqrt{-K}$ (by Enneper theorem ) have been beautifully integrated together by German engineer Otto Mohr and represented in his tensor diagram:
$$ k_n = k_1 \cos^2 \psi + k_2 \sin^2 \psi; \tau_g = (k_1 - k_2) \sin \psi \cos \psi; $$

The above is when reference axis coincides with the principal directions. Along principal directions $k_n=0$ else it has has extra term.
$$ k_n = k_1 \cos^2 \psi + k_2 \sin^2 \psi + k_{12}\sin \psi \cos \psi; $$
This is also explained in Mohr Circles Wiki.
Note that the radius makes $2 \psi$ to principal axis. Note also that the Gauss curvature changes depending on where we draw the $k_n$ line in the Mohr diagram.
Maximum geodesic torsion occurs when $\psi= \pi/4$when the line has a tendency to get away from the plane.
Special case of third kind is for  Cylindrical Parabolic/Flat  surfaces with $K=0$, that are developable, cones, cylinders. Practically seen in rolled paper, thin sheets etc.
You can appreciate how the curvature $ \tau_g $ brought in here explains the cross derivative curvature of a surface. It is torsion of all geodesics running through a point.
When $K<0$, real directions exist where $k_n$ vanishes. They are referred to as asymptotic directions. There two such  directions.
Practically we identify these direction using a straight edge of a ruler. When the edges cannot any more rotate in tangent plane around an axis normal to the surface axis then that is the asymptotic direction.
A: I like gt6989b answer. To elaborate a little bit, consider the 2 functions $f(x,y)=\frac{x^2+y^2}{2}, g(x,y)=xy$. Here's a cut section of these two functions, looking facing into the $y$ axis (cyan is $f$, magenta is $g$):

For $f$: $\frac{\partial f}{\partial x} = x$, and $\frac{\partial^2 f}{\partial x^2} = 1$. (And the same for $y$). But since $\frac{\partial f}{\partial x} = x$ is the same for every $y$, it is constant in $y$, and therefor $\frac{\partial^2 f}{\partial y\partial x}=0$.
For $g$ it's the opposite. If we first look at the partial on $x$, we can already see in the drawing from the straight lines, we'd get different value for each $y$. So $\frac{\partial^2 g}{\partial y\partial x}=1$, while $\frac{\partial^2 g}{\partial x^2} = 0$.
