We can achieve a simplified version of the Lorentz force by $$F=q\bigg[-\nabla(\phi-\mathbf{A}\cdot\mathbf{v})-\frac{d\mathbf{A}}{dt}\bigg],$$ where $\mathbf{A}$ is the magnetic vector potential and the scalar $\phi$ the electrostatic potential.
How is this derivable from a velocity-dependent potential $$U=q\phi-q\mathbf{A}\cdot\mathbf{v}?$$
I fail to see how the total derivative of $\mathbf{A}$ can be disposed of and the signs partially reversed. I'm obviously missing something.
Velocity-dependent potential is not strictly a potential. Lagrange equations say that
$$\frac{d}{dt}\frac{\partial L}{\partial \bf{v}} = \frac{\partial L}{\partial \bf{r}}$$
You have $L = L_0 - U$ where $L_0$ corresponds to free motion (e.g. $L_0 = mv^2/2$ or $L_0 = -mc^2\sqrt{1-(v/c)^2}$).
If $U$ does not contain $\bf{v}$ you have ${\partial L}/{\partial \bf{v}} = \bf{p}$ and so $\dot{\bf{p}} = -\nabla U$.
In this case, however $U$ contains $\bf{v}$ so on the you have
$$\frac{d}{dt}\left({\bf{p}} + q\bf{A}\right) = -\nabla U$$
Hints: Use
$$\frac{\partial U}{\partial {\bf v}}= -q{\bf A}, $$
and the defining property of a velocity-dependent potential:
$${\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}}.$$
See e.g. Herbert Goldstein, Classical Mechanics and Wikipedia for more details.
