Is $ f(x):= \frac{x}{ |x|^n }$ a gradient field? for $ 2 \leq n $ let be 
$ f: \mathbb{R}^n \backslash \{ 0 \} \rightarrow \mathbb{R}^n $
$ f(x):= \frac{x}{|x|^n} $ ,      $ x\in \mathbb{R}^n \backslash \{ 0 \} $
Is f a gradient field?
A vectorfield $ f $  ist a gradient field, if there is a function
$ f= \nabla v $
$ \left[ \nabla f (x) = (\frac{ \partial f}{ \partial x_1} (x),...,\frac{ \partial f}{ \partial x_n} (x)) \right] $
So, I need to find a function $v$ so that $ f= \nabla v $ right?
If I look for the case $n=2 $
so $ f(x):= \frac{x}{|x|^2} \in \mathbb{R}^n \backslash \{ 0 \} $
let be $ \partial_j v(x)= \frac{x_j}{|x|}$
then $v(x)= ln(|x|) $ and 
$$ \partial_j v(x)= \frac{1}{|x|} \frac{x_j}{|x|} = \frac{x_j}{|x^2|} $$
Is this the right idea?
If yes, how can I proceed for larger  $n$?
If no, what would you suggest?
Appreciate any help !
 A: Let's suppose that there exists some $V(x)$ such that $\frac{d}{dx}V(x)=f(x)$. 
Therefore:
$$\int_a^y\frac{d}{dx}V(x)dx =\int_a^yf(x) dx$$
Applying the fundamental theorem of calculus gives:
$$V(y)=\int_a^y \frac{x}{|x|^n} dx$$
Now let's break into cases. Suppose we are only considering $x,a,y > 0$. Then, we are left with a really simple integral, namely:
$$V(y)=\int_a^y \frac{x}{x^n} dx = \int_a^y \frac{1}{x^{n-1}} dx.$$
Now, let's assume we are only considering $x, a, y < 0$. This also simplifies very nicely:
$$V(y)=\int_a^y \frac{x}{(-x)^n} dx = (-1)^n\int_a^y \frac{1}{x^{n-1}} dx.$$
So it seems that we have a definition for $V$.
\begin{cases} 
      V(x) = \frac{1}{2-n}x^{2-n} & x < 0 \\
      V(x) = \frac{1}{2-n}(-1)^n x^{2-n} & x>0 
   \end{cases}
A: Your $f$ is a radial field and it depends only on the magnitude of your vector $r = |\vec{r}|$ so it is of the form $$f(\vec{r}) = \frac{\vec{r}}{r^n} = F(r)\hat{r}$$ where $F(r) = \frac1{r^{n-1}}$ and $\hat{r}$ is the unit radial vector.
Notice that for a scalar function $\phi : \mathbb{R}^n \to \mathbb{R}$ depending only on magnitude $r$ we have $$\nabla \phi(r) = \phi'(r)\nabla r = \phi'(r)\hat{r}$$
since $\nabla r = \hat{r}$.
Hence if $$\phi(r) = \int F(r)\,dr = -\frac1{(n-2)r^{n-2}}$$
then $f = \nabla \phi$.
