Differential of Lagrangian My professor wrote this $\frac{\partial L}{\partial q}\dot{q}=\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}})$. Due to the fact that I am very very very very bad at Math, could you explain me about this please? I don't understand.
$L(q,\dot{q},t)$ is the Lagrangian.
 A: First, the problem you are trying to solve is:
$$
S = \int\limits_{t_1}^{t_2} \mathcal{L}(q, \dot{q}, t)dt
$$
Assume there is a particular function, $q_0(t)$, which minimizes* this integral.  We perturb this function and find the value of the integral with the perturbed functions:
$$
q(\eta, t) = q_0(t) + \eta \lambda(t)
$$
All that we require is that $\lambda(t_1) = \lambda(t_2) = 0$ (this function vanishes at the end points of the integral).  This gives:
$$
S(\eta) = \int\limits_{t_1}^{t_2} \mathcal{L}(q_0 + \eta\lambda, \dot{q}_0 + \eta\dot{\lambda}, t)dt
$$
The tricky part (for me) is understanding that you can push the differentiation of $S$ inside of the integral:
$$
\frac{dS}{d\eta} = \int\limits_{t_1}^{t_2} \frac{d\mathcal{L}(q_0 + \eta\lambda, \dot{q}_0 + \eta\dot{\lambda}, t)}{d\eta}dt = 0 \\
\frac{dq}{d\eta} = \lambda, \frac{d\dot{q}}{d\eta} = \dot{\lambda}, \frac{dt}{d\eta} = 0
$$
\begin{align*}
\frac{d\mathcal{L}}{d\eta} =&\ \frac{\partial \mathcal{L}}{\partial q}\frac{dq}{d\eta} + \frac{\partial \mathcal{L}}{\partial \dot{q}}\frac{d\dot{q}}{d\eta} + \frac{\partial\mathcal{L}}{\partial t}\frac{dt}{d\eta}\\
 =&\ \frac{\partial \mathcal{L}}{\partial q}\lambda + \frac{\partial \mathcal{L}}{\partial \dot{q}}\dot{\lambda} = 0
\end{align*}
$$
\int\limits_{t_1}^{t_2}\left(\frac{\partial \mathcal{L}}{\partial q}\lambda + \frac{\partial \mathcal{L}}{\partial \dot{q}}\dot{\lambda}\right)dt = 0 \\
 \int\limits_{t_1}^{t_2}\frac{\partial \mathcal{L}}{\partial \dot{q}}\dot{\lambda}dt = -\int\limits_{t_1}^{t_2}\frac{\partial \mathcal{L}}{\partial q}\lambda dt
$$
Do integration by parts (i.e. the product rule) for the left integral:
\begin{align}
u =& \frac{\partial \mathcal{L}}{\partial \dot{q}} & du =& \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}}\right)dt \\
dv =& \dot{\lambda}dt & v =& \lambda 
\end{align}
$$
\int\limits_{t_1}^{t_2} \frac{\partial \mathcal{L}}{\partial \dot{q}}\dot{\lambda}dt = \left.\lambda \frac{\partial \mathcal{L}}{\partial \dot{q}}\right|_{t_1}^{t_2} - \int\limits_{t_1}^{t_2}  \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}}\right) \lambda dt
$$
But we said $\lambda(t_1) = \lambda(t_2) = 0$ so the first part of this integral vanishes, leaving:
$$
\int\limits_{t_1}^{t_2} \frac{\partial \mathcal{L}}{\partial \dot{q}}\dot{\lambda}dt = - \int\limits_{t_1}^{t_2}  \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}}\right) \lambda dt
$$
Substitute this back into the equation $\int\frac{\partial \mathcal{L}}{\partial \dot{q}}\dot{\lambda}dt = -\int\limits\frac{\partial \mathcal{L}}{\partial q}\lambda dt$
$$
- \int\limits_{t_1}^{t_2}  \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}}\right) \lambda dt =  -\int\limits_{t_1}^{t_2}\frac{\partial \mathcal{L}}{\partial q}\lambda dt \\
\int\limits_{t_1}^{t_2} \left(\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}}\right) - \frac{\partial \mathcal{L}}{\partial q}\right)\lambda dt = 0
$$
For this integral to be zero for all functions $\lambda$, the thing multiplying $\lambda$ must be zero, hence the Euler-Lagrange equation:
$$
\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}}\right) - \frac{\partial \mathcal{L}}{\partial q} = 0
$$
*I say "minimize"--this is not correct. It's not even correct to say "find extremum". In reality, we're simply finding a function that is analogous to a critical point. And likewise, once we find such a function, we must verify it's the extremum we're looking for.
