electromagnetism - Deriving the Lorentz force from velocity dependent potential

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.

Answer

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.