Clairaut's theorem

Mixed partial derivatives of $C^2$-functions are equal.

Generally, given a function $$f \colon {\bf R}^n \to {\bf R},$$ which has continuous second partial derivatives at a given point $(a_1, \dots, a_n) \in {\bf R}^n$, then for $1 \leq i,j \leq n,$, we have $$\frac{\partial^2 f}{\partial x_i\, \partial x_j}(a_1, \dots, a_n) = \frac{\partial^2 f}{\partial x_j\, \partial x_i}(a_1, \dots, a_n).\,$$

So, the operations of partial differentiation commute: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} C^2 & \ra{\frac{\partial}{\partial x}} & C^1 \\ \da{\frac{\partial}{\partial y}} & \searrow & \da{\frac{\partial}{\partial y}} \\ C^1 & \ra{\frac{\partial}{\partial x}} & C \end{array} $$