A common derivation of the Lagrangian of a charged particle in an electromagnetic field starts with the Lorentz force that is rewritten in terms of the electromagnetic potentials $\Phi(\vec{x})$ and $\vec{A}(\vec{x}, t)$. So $\vec{F} = q(\vec{E} + \frac{1}{c}\dot{\vec{x}}\wedge\vec{B})$ becomes
\begin{align} \vec{F} & = q(-\nabla\Phi-\frac{1}{c}\frac{d\vec{A}}{dt} + \frac{1}{c}\nabla\cdot(\dot{\vec{x}}\cdot\vec{A})) \notag \\ & = \nabla(-q\Phi+\frac{q}{c}(\dot{\vec{x}}\cdot\vec{A})) - \frac{q}{c}\frac{d\vec{A}}{dt}, \label{u} \end{align}
where $q$ is the charge of the particle. The Lagrangian takes the general form $L = T-V$. In this case the kinetic energy is simply $T = \frac{1}{2}m\dot{\vec{x}}^{2}$, where $m$ is the mass of the particle. For the potential energy one sets $V = q(-\Phi+\frac{1}{c}(\dot{\vec{x}}\cdot\vec{A})).$ By the look of the final equation for $\vec{F}$ \eqref{u} this seems like a fairly reasonable choice, except for the fact that the term $- \frac{q}{c}\frac{d\vec{A}}{dt}$ is omitted. I'm aware that this term conveniently reemerges when one computes the Euler-Lagrange equations. I also know that two Lagrangians that only differ by a total time derivative yield the same Euler-Lagrange equations. However, I do wonder whether there's a reason for omitting this term in the definition of $V$. I'm not suggesting that we simply add it to $V$, that would likely be nonsensical. I just want to understand what motivates the omission because as it stands, it seems like this step is pulled out of thin air.
