I have seen that in every derivation Starting from $L= \frac{1}{2}m\dot{x}^{2} - e\dot{x}\cdot\textbf{A}$ the equation $F= \frac{1}{2}mv^2+q(\vec{v} \times \vec{B} )$ is derived. can we derive Lagrangian $L= \frac{1}{2}m\dot{x}^{2} - e\dot{x}\cdot\textbf{A}$? starting from force? Euler-Lagrangian equation relates force to the Lagrangian,I'm having trouble in this part $ e\dot{x}\cdot\textbf{A}$. Ho to get this part?
After substituting $B=\nabla \times \vec{A} $ and substituting identity $\vec{v} \times (\nabla \times \vec{A}) = \nabla(\vec{v}\cdot\vec{A})-(\vec{v}\cdot\nabla)\vec{A} $ I'm having trouble in the simplification of magnetic force. $q(\vec{v}\times \vec{B}) = q[\vec{v} \times (\nabla \times \vec{A})] = q[\nabla(\vec{v}\cdot\vec{A})-(\vec{v}\cdot\nabla)\vec{A}]$ Then how to simplify further? In the final expression of Lagrangian there is no $\nabla$ so how to get rid of this?