Two different formulas My problem is simple : given a particle of mass $m$, charge $q$ and velocity $\bf{v}$. If $\bf{A}$ denotes the magnetic potential satisfying $\bf{B}= \nabla \times \bf{A}$. 
I want to etablish the laws of motion in the theory of special relativity for this particle. 
And the problem is that I need to compute $\nabla (\bf{v} \cdot \bf{A})$. 
I read on several textbooks that : $$\nabla (\bf{v} \cdot \bf{A}) = \bf{v} \times (\nabla \times \bf{A}) + (\bf{v} \cdot \nabla) \bf{A} $$
The problem is that the general formula according to wikipedia is $$\nabla (\bf{v} \cdot \bf{A}) = \bf{v} \times (\nabla \times \bf{A}) + \bf{A} \times (\nabla \times \bf{v})+ (\bf{v} \cdot \nabla) \bf{A} + (\bf{A} \cdot \nabla) \bf{v}  $$
[with general vectors $\bf{v}$ and $\bf{A}$]
So I don't understand why $\bf{A} \times (\nabla \times \bf{v})+(\bf{A} \cdot \nabla) \bf{v} =0$ if one does not make any other assumption over $\bf{v}$ and $\bf{A}$. 
I think people here have encouter this prolem so please if you walk by and know, just give me a hint of a link.  
 A: If you look at the relativistic lagrangian:
$$\mathcal{L} = -m_{0}c^2\sqrt{1 - \frac{v^2}{c^2}} + q \vec{A} \cdot \vec{v}$$
Equations of motion are derived from:
$$\frac{\partial \mathcal{L}}{\partial x_{i}} - \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{x}_{i}} = 0$$
In our case:
$$\frac{\partial \mathcal{L}}{\partial x_{i}} = \frac{\partial }{\partial x_{i}}(\vec{A} \cdot \vec{v}) = \vec{v} \cdot \frac{\partial \vec{A}}{\partial x_{i}}$$
We do not differentiate over $\vec{v} = \dot{\vec{x}}$, cause it is an independent variable in Lagrange formalism.
More generally:
$$\nabla (\vec{A} \cdot \vec{v})$$
and it only acts on magnetic potential $\vec{A}$, so we can write (summation convention):
$$\begin{align}
\left[ \nabla (\vec{A} \cdot \vec{v})\right]_{i} & = v_{j} \nabla_{i} A_{j} \\[3mm]
& = v_{j} \nabla_{i} A_{j} - v_{j}\nabla_{j}A_{i} + v_{j}\nabla_{j}A_{i} \\[3mm]
& = \left(\delta_{i l} \delta_{jm} - \delta_{im}\delta_{jl} \right) v_{j} \nabla_{l} A_{m} + v_{j}\nabla_{j}A_{i} \\[3mm]
& = \epsilon_{ijk}\epsilon_{klm}v_{j}\nabla_{l}A_{m} + v_{j}\nabla_{j}A_{i} \\[3mm]
& = \left[ \vec{v} \times (\nabla \times \vec{A}) + (\vec{v} \cdot \nabla) \vec{A} \right]_{i}
\end{align}$$
A: In Lagrangian formalism you have two types of variables $q = q(t)$ and $\dot q = \dot q(t)$. These variables are functions of time (this implies $\partial_{i}q = 0$ and $\partial_{i}\dot q = 0 \text{ for } i = \{1, 2, 3\}$). This is how Lagrangian works, otherwise one would have something really complicated
From Nex_Friedrich's answer follows that $q := \vec x$ and $\dot q: = \dot{ \vec x} = \vec v $. Thus you see that in our case $\nabla \dot q = \nabla \vec v = 0$.
That's how you can justify rule from wikipedia. 
