Classical Mechanics: Relation between general velocity and general potential function for velocity-dependent potential

Question

How is the general force derived from the general potential for a velocity-dependent potential $U = U(q_j,\dot{q_j})$?
$$Q_j=-\frac{\partial U}{\partial q_j}+ \frac{\mathrm{d}}{\mathrm{dt}}(\frac{\partial U}{\partial \dot{q_j}}). \tag{1}$$

What I understand

1. In a conservative vector field, the force could be found through the gradient of the scalar potential, i.e. $F = -\nabla V$.
2. General force in non-velocity-dependent potentials is given by $Q_j = -\frac{\partial V}{\partial q_j}$, in a similar manner as of that in bullet point 1, where $V$ is a conservative vector field, $Q_j$ is the general force and $q_j$ a general coordinate.

My attempt to derive Eqn. (1)

1. Recall that the general potential is a function of general coordinate and the general velocity, $U = U(q_j,\dot{q_j})$

2. Apply $Q_j = -\nabla U$, and get
   $$
   \begin{aligned}
   Q_j &= -\nabla U(q_j,\dot{q_j})\\
   &=-(\frac{\partial}{\partial q_j}+\frac{\partial}{\partial \dot{q_j}})U(q_j,\dot q_j)\\
   &=-\frac{\partial U}{\partial q_j}-\frac{\partial U}{\partial \dot q_j}
   \end{aligned} \tag{2}
   $$
   where the second term on R.H.S. does not match Eqn. (1)

My question

1. What are the things/concepts that I am missing?
2. How is Eqn.(1) derived?
3. How to intuitively/physically understand the second term on R.H.S. in Eqn.(1)?

Answer 1

• 1 & 2: Eq. (1) is not derived. It is a defining property of a generalized velocity-dependent potential.

• 3. The form of eq. (1) mimics the Euler-Lagrange operator, so that we can bring Lagrange equations $$ \frac{d}{dt}\frac{\partial T}{\partial \dot{q}^j}-\frac{\partial T}{\partial q^j}~=~Q_j, \qquad j~\in \{1,\ldots, n\}, \tag{L}$$ in the form of Euler-Lagrange equations $$ \frac{d}{dt}\frac{\partial (T-U)}{\partial \dot{q}^j}-\frac{\partial (T-U)}{\partial q^j}~=~0, \qquad j~\in \{1,\ldots, n\}. \tag{EL}$$