Lagrangian gauge invariance $L'=L+\frac{df(q,t)}{dt}$ So, I have to prove directly (e.g. by substitution) that if a path satisfies the Euler-Lagrange equations for the Lagrangian $L$ it does so for $$L'=L+\frac{df(q,t)}{dt}.$$ Let me tell you what I have done:
$$\frac{d}{dt}\frac{\partial L'}{\partial \dot{q_i}}-\frac{\partial L'}{\partial q_i}=(\frac{d}{dt}\frac{\partial L}{\partial \dot{q_i}}-\frac{\partial L}{\partial q_i})+(\frac{d}{dt}\frac{\partial}{\partial \dot{q_i}}\frac{df}{dt}-\frac{\partial}{\partial q_i}\frac{df}{dt})$$
I would like to somehow prove that the second parenthesis is $0$ to this end I have used the following:
$$\frac{\partial}{\partial \dot{q_i}}=\sum\frac{\partial q_j}{\partial \dot{q_i}}\frac{\partial}{\partial q_j} \Rightarrow \frac{d}{dt}\frac{\partial}{\partial \dot{q_i}}=\sum \frac{d}{dt}\frac{\partial q_j}{\partial \dot{q_i}}\frac{\partial}{\partial q_j}=\sum \frac{\partial \dot{q_j}}{\partial \dot{q_i}}\frac{\partial}{\partial q_j}.$$
Now the proof is complete IF $$\frac{\partial \dot{q_j}}{\partial \dot{q_i}}=δ_{ij} .$$ I know this is true for $q_i,q_j$ but is it also true for their derivates?
 A: Simplest is to go with
\begin{align}
\frac{df}{dt}&= \frac{\partial f}{\partial q}\dot q + \frac{\partial f}{\partial t}\, ,\\
\frac{d}{dt}\left(\frac{\partial }{\partial \dot{q}}\frac{df}{dt}\right)&= \frac{d}{dt}\left(\frac{\partial f}{\partial q} \right)\\
&= \frac{\partial }{\partial q}\left(
\frac{df}{dt}\right)
\end{align}
You can deal with the multivariable case yourself.
A: Just for completeness, another more straightforward way to do this is by considering that by solving the Euler-Lagrange equations you are essentially finding a stationary point of the action $S = \int_{t_i}^{t_f} L dt$. In other words, you find the path $q(t)$ which minimizes the action, given fixed end point $q(t_f)$, $q(t_i)$. This is the procedure that yields the Euler-Lagrange equations.
Having said that, consider then the action $S' = \int_{t_i}^{t_f} dt L' = \int_{t_i}^{t_f} dt\left[ L +\frac{df}{dt}\right]$
Then, integrating :
$$ S' = \int_{t_i}^{t_f} dt L + [f(q,t)]^{t_f}_{t_i} = S+f(q(t_f),t_f)-f(q(t_i),t_i)$$
Now, the end points of the trajectory are fixed while finding the stationary point. Hence, when we variate $S$, we variate the path, but not the end points. In other words :
$$\delta S' = \delta S + \underbrace{\delta\big(f(q(t_f),t_f)-f(q(t_i),t_i)\big)}_{= 0} \Leftrightarrow \delta S' = \delta S$$
Hence, the variation of $S'$ and $S$ is the same, so they provide equivalent equation of motion !
A: You're on the right track and almost there with your proof, you need to look at something called 'cancelation of dots' a useful little trick when doing Lagrangian Mechanics. A good proof is given here  by Bernhard Heijstek.
