I was reading the solution of this exercise and I have a doubt:
A point particle moves in space under the influence of a force derivable from a generalized potential of the form $$U(r,v) = V(r)+\sigma\cdot L $$
where $r$ is the radius vector from a fixed point, $L$ is the angular momentum about that point, and $\sigma$ is a fixed vector in space.
I need to find the components of the force on the particle in Cartesians coordinates, on the basis of Lagrange equations with a generalized potential.
This exercise is from the book "Goldstein - Classical Mechanics". I have the solution, but I really don't understand a step. If I convert $r$ to Cartesian coordenates, I have $r = \sqrt{x^2+y^2+z^2}$, and if I put the expression into the Lagrange equation: $$Q_{j} = \frac{d}{dt}\frac{\partial }{\partial \dot q_{j}}\left(V\left(\sqrt{x^2+y^2+z^2}\right)+\sigma\cdot L\right) - \frac{\partial }{\partial q_{j}}\left(V\left(\sqrt{x^2+y^2+z^2}\right)+\sigma\cdot L\right) $$
$$= \frac{d}{dt}\frac{\partial}{\partial \dot v_{j}}(\sigma\cdot[(x\hat{i}+y\hat{j}+z\hat{k})\times(p_{x}\hat{i}+p_{y}\hat{j}+p_{z}\hat{k})])- \frac{\partial }{\partial x_{j}}\left(V\left(\sqrt{x^2+y^2+z^2}\right)+\sigma\cdot L\right)$$
What happened with $V\left(\sqrt{x^2+y^2+z^2}\right)$? and why $\partial\dot q_{j} = \partial\dot v_{j}$?
If anyone can give me an explanation I would appreciate it :)