Is there an intuitive explanation for the equality of mixed partials? The fact that the mixed second order partial derivatives of a $C^2$ smooth scalar valued function are equal seems, to me, quite surprising. For example, if you interpret $\frac{\partial ^2f}{\partial y \partial x}$ as the change of the slope of a tangent line along the $x$-axis when moving along the $y$-axis, it's not at all obvious to me that this should equal $\frac{\partial ^2f}{\partial x \partial y}$.
Does there exist some intuitive or visual way to explain this equality? Just to be clear, I'm not looking for a formal proof (this can be found in most textbooks), but some intuitive reason or hint as to why this might be true. 
 A: One intuition is that if the function is smooth enough then it is well approximated by its second degree Taylor polynomial, and equality of mixed partials is straightforward for those polynomials.
Let's focus on just two variables; nothing really changes if there are more. An arbitrary quadratic in $x$ and $y$ looks like
$$
q(x,y) = a + bx + cy + dx^2 + exy + fy^2
$$
If we compute $(\partial^2 q)/(\partial x \partial y)$ we obtain $e$, and if we compute $(\partial^2 q)/(\partial y \partial x)$ we also obtain $e$. So in this special case the mixed partials are equal.
If $f(x,y)$ is smooth enough, it is well approximated by its Taylor polynomials. For example, the first-degree Taylor polynomial is the equation of the tangent plane at a given point, and this first-degree polynomial agrees with $f$ on the first partial derivatives.  If $f$ is smooth enough, the second-degree Taylor polynomial gives a quadratic function in $x$ and $y$ that agrees with the first and second derivatives. In particular, this means the mixed second partials of $f$ will be equal.  
In the special case where we have a second-degree Taylor polynomial of a function $f(x,y)$ at the point $(0,0)$, the polynomial has this form:
$$
\begin{split}
p(x,y) &= f(0,0) + \frac{\partial f}{\partial x}\Big |_{(0,0)}\cdot x +\frac{\partial f}{\partial y}\Big |_{(0,0)}\cdot y\\
& + \frac{\partial^2 f}{\partial x^2}\Big |_{(0,0)}\cdot \frac{x^2}{2}+ \frac{\partial^2 f}{\partial x\partial y}\Big |_{(0,0)}\cdot xy + \frac{\partial^2 f}{\partial y^2}\Big |_{(0,0)}\cdot \frac{y^2}{2}    
\end{split}
$$
