A vector field is spherically symmetric about the origin if, on every sphere centered at the origin, it has constant magnitude and points either away from or toward the origin. A vector field that is spherically symmetric about the origin can be written in terms of the spherical coordinate $F = f(\rho)e_\rho$ where $f$ is a function of the distance $ρ$ from the origin, $f(0) = 0$, and $e_ρ$ is a unit vector pointing away from the origin.
Show that $\operatorname{div} F = \frac 1 {\rho^2} \frac{d(\rho^2f(\rho))}{d \rho} , ρ \neq 0.$
Source: Hughes-Hallet, freely available here (project #2).
My work is below. I request verification or feedback.
Note: This problem is trivial to solve using the Divergence Theorem in Spherical Coordinates. But that has not yet been taught in the book; the point of this problem is to derive Divergence Theorem in Spherical Coordinates, or at least as much as you need to solve the problem.
Consider the point $v_0$ on the $x$ axis such that $$\begin{align*}v_0 &= \langle \rho_0, 0, 0 \rangle\\ f_0 &= f(\rho_0)\\ F(v_0) &= \frac {f_0}{\rho_0}v_0.\end{align*}$$ Taking a step $\partial x$ in the $x$ direction doesn't change the direction of $v$, but only increases $\rho$, giving $$\begin{align*} v_1 &= \langle \rho_0 + \partial x, 0, 0 \rangle \\ f_1 &= f(\rho_0 + \partial x) \\ F(v_1) &= \frac {f_1}{\rho_0}v_0 \\ \frac{\partial F_x}{\partial x} &= \frac{\partial f}{\partial \rho}. \end{align*}$$
Now, instead of stepping in the $x$ direction, consider starting from $v_0$ and taking a step $\partial y$ in the $y$ direction: $$v_2 = \langle \rho_0, \partial y, 0 \rangle.$$ Since this step is orthogonal to $v_0$, $$\rho_2 = \rho_0 + \mathcal{O}([\partial y]^2)$$ and since $f$ depends only on $\rho$, we have to a first order approximation $$f_2 \approx f_0 \\ F(v_2) \approx \frac {f_0}{\rho_0}v_2$$ and, in the limit, $$\frac{\partial F_y}{\partial y} = \frac f \rho.$$
By symmetry, $\frac{\partial F_z}{\partial z} = \frac f \rho$ as well, giving $$\begin{align*} \operatorname{div} F &= \frac{\partial f}{\partial \rho} + 2\frac f \rho \\ &= \frac 1 {\rho^2} \frac{d(\rho^2f(\rho))}{d \rho} \end{align*}$$ as desired.
The above applies directly to points on the $x$ axis (other than the origin), and, by symmetry to any point in space (besides the origin).