consider a particle in a central potential, i.e. the potential $V$ only depends on the distance $ r = \| \vec{x} \|$ to the origin. The equation of motion thus reads
$$ m\ddot{\vec{x}}=-\frac{\partial}{\partial \vec{x}}V(\vec{x})=-\frac{\vec{x}}{r}\frac{dV(r)}{dr} \tag{1}$$
We also know that the angular momentum is conserved:
$\frac{d}{dt} \vec{L} = \dot{\vec{x}}\times m\dot{\vec{x}} + \vec{x}\times m\ddot{\vec{x}} \tag{2}=0$
since $\dot{\vec{x}} \parallel \dot{\vec{x}}$ and $\vec{x} \parallel \vec{F}$. The later is true because we are in a central potential.
Now we demand that the Laplace-Runge-Lenz vector $\vec{A}$ is conserved. We want to know a condition for that to hold.
We have
$\vec{A} = \dot{\vec{x}} \times \vec{L} + V(r)\vec{x} \tag{3}$
so
$\begin{align} \frac{d}{dt}\vec{A} &=\ddot{\vec{x}}\times L + \frac{1}{r}\frac{dV(r)}{dr}(\vec{x}\cdot\dot{\vec{x}})\vec{x} + V(r)\dot{\vec{x}} \tag{4} \\ &=-\frac{1}{r}\frac{dV(r)}{dr}\vec{x} \times ( \vec{x}\times \dot{\vec{x}}) + \frac{1}{r}\frac{dV(r)}{dr}(\vec{x}\cdot\dot{\vec{x}})\vec{x} + V(r)\dot{\vec{x}}\\ &= \big[r\frac{dV(r)}{dr} + V(r)\big] \stackrel{!}{=}0 \end{align}$
Now my problem is basically understanding
$ \frac{d}{dt} V(r)\vec{x} \tag{5}$
I get
$ \begin{align} \frac{d}{dt} V(r)\vec{x} & = \frac{d}{dt} V(\|x\|)\vec{x}\\ &= \frac{d \|x\|}{dt}\frac{dV(\|x\|)}{dt}\vec{x} + V(\|x\|)\dot{\vec{x}}\\ &= \frac{\vec{x}}{r}\frac{dV(r)}{dr}\vec{x} + V(\|x\|)\dot{\vec{x}}\\ &= r\frac{dV(r)}{dr} + V(\|x\|)\dot{\vec{x}} \end{align}\tag{6}$
I'm unsure if I can replace $\frac{d}{dt}$ with $\frac{d}{dr}$ the way Idid it.
Also while it might look good it doesn't seem to be correct. Because if we look at (4) and (6) we end up with the same but started at two different points (not literally). One time we derived the Laplace-Runge-Lenz vector and the other time we derived the potential.
Where's my problem?
