with$$ x\in \mathbb{R} \setminus \{0\} $$ I am not sure how to approach this problem. div is defined as the following, where F is an differentiable vector function: $$\operatorname{div} F:=\sum_{i=1}^{n} \partial_{i} F_{i}$$
I thought about defining $$u: x \mapsto \frac{1}{||x||_{2}}$$ and then using, $$\operatorname{div}(u F)=\langle\operatorname{grad} u, F\rangle+u \operatorname{div} F$$ but I don't see how x can be an differentiable vector function. I'd be very thankful for an approach.
First, definitions: Let $\mathbf{X}=(x_1,...,x_n)$, let $\Vert\mathbf{X}\Vert=\sqrt{\sum_{i=1}^{n}{{x_i}^2}}$, and let $\hat{\mathbf{X}}=\frac{\mathbf{X}}{\Vert\mathbf{X}\Vert}$. We wish to show that $$\nabla \boldsymbol{\cdot} \hat{\mathbf{X}}=\frac{n-1}{\Vert\mathbf{X}\Vert}.$$ Well, using the quotient rule, it's clear that $$\frac{\partial}{\partial x_i} \frac{x_i}{\Vert\mathbf{X}\Vert} = \frac{\Vert\mathbf{X}\Vert^2-{x_i}^2}{\Vert\mathbf{X}\Vert^3}$$ Therefore, $$\nabla \boldsymbol{\cdot} \hat{\mathbf{X}} = \sum_{i=1}^{n}\frac{\partial}{\partial x_i} \frac{x_i}{\Vert\mathbf{X}\Vert}$$
$$=\sum_{i=1}^{n}{\frac{\Vert\mathbf{X}\Vert^2-{x_i}^2}{\Vert\mathbf{X}\Vert^3}}$$
$$=\frac{n-1}{\Vert\mathbf{X}\Vert}.$$ Feel free to ask about anything you don't understand!