[proofplan] We expand the squared Euclidean norm in coordinates and compute the coordinate partial derivatives directly. Each first [partial derivative](/page/Partial%20Derivative) is the linear function $2x_i$, so each second derivative in the same coordinate direction is the constant $2$. The Euclidean Laplacian is the sum of these $m$ second coordinate derivatives, giving $2m$ at every point. [/proofplan] [step:Expand the squared Euclidean norm in coordinates] Let $x=(x_1,\ldots,x_m)\in\mathbb{R}^m$. By the definition of the Euclidean norm on $\mathbb{R}^m$, \begin{align*} u(x)=|x|^2=\sum_{j=1}^m x_j^2. \end{align*} Thus $u$ is a polynomial function on $\mathbb{R}^m$, so $u$ is twice continuously differentiable. [/step] [step:Compute each second coordinate partial derivative] Fix $i\in\{1,\ldots,m\}$. Holding all coordinates except $x_i$ fixed, the expression \begin{align*} u(x)=\sum_{j=1}^m x_j^2 \end{align*} has only one term depending non-constantly on $x_i$, namely $x_i^2$. Therefore \begin{align*} \partial_{x_i}u(x)=2x_i. \end{align*} Differentiating once more in the same coordinate direction gives \begin{align*} \partial_{x_i}(\partial_{x_i}u)(x)=2. \end{align*} [guided] Fix $i\in\{1,\ldots,m\}$. We compute the partial derivative in the $x_i$ direction by treating the remaining coordinates $x_j$ with $j\ne i$ as constants. Since \begin{align*} u(x)=\sum_{j=1}^m x_j^2, \end{align*} the term $x_i^2$ contributes derivative $2x_i$, while every term $x_j^2$ with $j\ne i$ contributes derivative $0$ with respect to $x_i$. Hence \begin{align*} \partial_{x_i}u(x)=2x_i. \end{align*} Now differentiate this first partial derivative again with respect to the same coordinate $x_i$. The map $x\mapsto 2x_i$ is linear in the $i$-th coordinate, so its $x_i$-partial derivative is the constant $2$. Thus \begin{align*} \partial_{x_i}(\partial_{x_i}u)(x)=2. \end{align*} This calculation is independent of the point $x$, which is why the final Laplacian will be constant on all of $\mathbb{R}^m$. [/guided] [/step] [step:Sum the second coordinate partial derivatives to obtain the Laplacian] By the coordinate definition of the Euclidean Laplacian on $\mathbb{R}^m$, \begin{align*} \Delta u(x)=\sum_{i=1}^m \partial_{x_i}(\partial_{x_i}u)(x). \end{align*} Using the computation from the previous step for each $i\in\{1,\ldots,m\}$, we obtain \begin{align*} \Delta u(x)=\sum_{i=1}^m 2=2m. \end{align*} Since $x\in\mathbb{R}^m$ was arbitrary, this holds for every $x\in\mathbb{R}^m$. [/step]