I have a problem proving the following:
Given is a function $u: \mathbb{R}^n \backslash 0 \rightarrow \mathbb{R}^n $ with $u(x)= x |x|^{-n}$
Now i need to show that for all $x \in \mathbb{R}^n \backslash 0$
$$\operatorname{div} u(x) = 0$$
To be honest i dont even know how to start. Can anyone help me to prove that? It seems to be simple but apparently im stuck at the moment
As pointed out in the comment you can directly apply the formula for the divergence: $$\operatorname{div} (u ) = \sum_{k=1}^n \frac{\partial u_k}{\partial x_k}$$ as $$u_k(x)=x_k |x|^{-n}=x_k \left(\sum_{j=1}^n x_j^2 \right)^{-\frac{n}{2}}$$ you obtain: $$\frac{\partial u_k}{\partial x_k}=|x|^{-n}-\frac{n}{2} x_k \times \frac{2 x_k}{\left(\sum_{j=1}^n x_j^2 \right)^{-\frac{n}{2}+1}}=\frac{1}{|x|^{n+2}} \left( |x|^2-n x_k^2\right) $$ summing over $k$: $$\operatorname{div}(u)=\sum_{k=1}^n\frac{1}{|x|^{n+2}} \left( |x|^2-n x_k^2\right) =\frac{1}{|x|^{n+2}} \left(n |x|^2- n \sum_{k=1}^n x_k^2\right)=0 $$
Remark
This straightforward approach is probably one of the least efficient, you can for example use the formula for the divergence of a radial function or a intrinsic approach.