Showing a vector field is not conservative Let $f: R^n \to R^n$. $||x||$ is Euclidean norm. Define $f(x) = g(||x||)$. where $g: [0, \infty) \to R^n$ is differentiable on $(0, \infty)$. I want to show that $f$ is not conservative on $R^n$. ($g$ is not constant)
MY Effort:
Write $f(x) = (f_1(x),...,f_n(x))$ and $g(||x||) = (g_1(||x||),...,g_n(||x||))$ Hence,
$$ f_i = g_i(||x||) $$. In particular
$$ \frac{ \partial f_i}{\partial x_j} = \frac{x_j}{\sqrt{x_1^2+...+x_n^2}} \cdot \frac{d g_i }{d x_j}(||x||)$$
and
$$ \frac{ \partial f_j}{\partial x_i} = \frac{x_i}{\sqrt{x_1^2+...+x_n^2}} \cdot \frac{d g_j }{d x_i}(||x||)$$
Is this enough to conclude that mixed partials are not equal, therefore $f$ is not conservative?
 A: We are given a vector-valued function
$${\bf g}:\qquad t\mapsto{\bf g}(t)=\bigl(g_1(t),\ldots,g_n(t)\bigr)\qquad(t>0)$$
and define a vector field ${\bf F}=(F_1,\ldots,F_n)$ on $\dot{\mathbb R}^n$ by
$${\bf F}({\bf x}):={\bf g}\bigl(|{\bf x}|\bigr)\ .$$
The integrability condition for this field reads
$$F_{i.j}({\bf x})-F_{j.i}({\bf x})\equiv 0\qquad \forall i\ne j\ .\tag{1}$$
Now according to the chain rule we have
$$F_{i.j}({\bf x})=g_i'\bigl(|{\bf x}|\bigr)\>{\partial|{\bf x}|\over\partial x_j}=g_i'\bigl(|{\bf x}|\bigr){x_j\over{|\bf x}|}\ .$$
The condition $(1)$ can therefore be interpreted as follows: The matrix
$$\left[\matrix{g_1'\bigl(|{\bf x}|\bigr)&g_2'\bigl(|{\bf x}|\bigr)&\ldots&g_n'\bigl(|{\bf x}|\bigr)\cr
x_1&x_2&\ldots&x_n\cr}\right]\tag{2}$$
must have all subdeterminants of order $2$ equal zero, in other words: have rank $\leq1$, for all ${\bf x}\in\dot{\mathbb R}^n$.
Assume now that ${\bf g}'(r)\ne{\bf 0}$ for some $r>0$. Then the rank condition on the matrix $(2)$ would imply that all vectors ${\bf x}$ on the sphere $S^{n-1}_r$ of radius $r$ are multiples of ${\bf g}'(r)$, which is absurd.
A: This is a comment with unwieldy formulas.
The correct formulas:
$$
\frac{ \partial f_i}{\partial x_j} = \frac{x_j}{\sqrt{x_1^2+...+x_n^2}} \cdot g_i'(||x||)
$$
($g$ is a  vector-valued 1-var function)
