Laplacian as a sum of second partial derivatives

Statement

Let $U \subseteq \mathbb{R}^n$ be open, and let $f \colon U \to \mathbb{R}$ be twice differentiable on $U$. Then $$\Delta f = \frac{\partial^2 f}{\partial x_1^2} + \frac{\partial^2 f}{\partial x_2^2} + \dots + \frac{\partial^2 f}{\partial x_n^2}.$$

Remarks

Proof: $\nabla f = (\partial f / \partial x_1, \dots, \partial f / \partial x_n)$, and the divergence adds the derivative of the $i$-th component with respect to $x_i$. So $\Delta f$ is the trace of the Hessian matrix of $f$. The trace does not change under a rotation of coordinates, so neither does $\Delta f$. Examples: $\Delta (x^2 + y^2) = 4$, $\Delta (x^2 - y^2) = 0$ and $\Delta \|x\|^2 = 2n$ on $\mathbb{R}^n$. In polar coordinates in the plane, $\Delta f = f_{rr} + \tfrac{1}{r} f_r + \tfrac{1}{r^2} f_{\theta\theta}$ for $r > 0$.

Explanations

L1 human

In Cartesian coordinates, the Laplacian adds up the second derivatives of $f$ in the $n$ coordinate directions. Only the unmixed second partial derivatives appear, not mixed ones like $f_{xy}$.