This question is related to Lagrangian gauge invariance and Adding a total time derivative term to the Lagrangian but now what if the function $F = F(\{q_i\}, \{\dot{q}_i\},t)$? Will the Lagrangian $L' = L + dF/dt$ still yield the same Euler-Lagrange equations?
My gut instinct says no, but I am not sure how to prove/show this. For the total differential of $F$ I get $$dF = \frac{\partial F}{\partial t}dt + \frac{\partial F}{\partial q}dq + \frac{\partial F}{\partial \dot{q}}d\dot{q},$$ but when I try to perform the same procedure as those linked above, I don't believe I find the correct derivatives when plugging it into the Euler-Lagrange equations.
Any help or ideas is appreciated. I am not sure if I have to go back to the action and perform the variation there to find the root cause or what. I know there are second-order Euler-Lagrange equations, but I don't think this is hinting at that.
For example, when I evaluate $\frac{\partial}{\partial\dot{q}}\left(\frac{dF}{dt}\right)$, I find that the $\partial F/\partial t$ term goes away, but I am left with $$\frac{\partial F}{\partial\dot{q}} + \frac{\partial}{\partial \dot{q}}\left(\frac{\partial F}{\partial\dot{q}}\ddot{q}\right)$$, which I do not know how to evaluate the last term, or just get rid of the second order derivative since no $\dot{q}$s are present. This may be trivial and I am over thinking. If it does go away, then I will be left with the same result when $\dot{q}$ is not present in $F$.
