I'm asked to compute the laplacian of $$ f: \mathbb{R}^n \setminus \{0\} \to \mathbb{R},\; x \mapsto |x|^\alpha $$
I know $$ \begin{align*} \Delta f(x) &= \nabla \cdot (\nabla f) \\ &= \nabla \cdot \begin{pmatrix}\alpha x_1 \left(f(x)\right)^{1 - \frac{2}{\alpha}} \\ \alpha x_2 (f(x))^{1 - \frac{2}{\alpha}} \\ \alpha x_3 \left(f(x)\right)^{1 - \frac{2}{\alpha}} \\ \vdots \end{pmatrix} \end{align*} $$
Calculating the laplacian for dimensions $n = 1 \ldots 4$ I noticed it to likely be $$ \Delta f(x) = \alpha (\alpha + (n-2)) \left( \sum_{i=1}^n x_i^2 \right)^{\alpha/2-1} $$
Am I supposed to just partially differentiate each entry of the column vector with respect to $x_i$ and add up all the second order partial derivatives? (i.e. $\frac{\partial^2 f(x)}{\partial x_i \partial x_i})$