pccrp said:
Since in the start of the demonstration the \frac{\partial L}{\partial q_j} treats \dot q_j as constants and vice versa, they'll keep this behavior on the Lagrange's equations.
Am I correct? Thanks for your help
Yes, this is the idea. However, I'm not happy with phrases like "consider as independent" and "treat as constants". This sounds like one could arbitrarily choose how to interpret the derivatives. That's not the case, though. You can prove the Euler-Lagrange equations with full rigour by strictly applying the rules of calculus. There is no freedom how to interpret terms.
Here is how I would derive the Euler-Lagrange equations (leaving out all technicalities for the sake of simplicity):
We want to find the trajectory ##q(t)## that makes the action
S[q] = \int_{t_a}^{t_b} L(q(t),\dot q(t))\mathrm d t
stationary. A necessary condition for this to be true is that whenever we add a multiple of some arbitrary function ##\eta(t)## with ##\eta(t_a) = \eta(t_b) = 0## to ##q(t)##, ##S[q+\epsilon\eta]## shouldn't change much for small ##\epsilon##. Since, given fixed ##q## and ##\eta##, ##S[q+\epsilon\eta]## is just a real-valued function of the real parameter ##\epsilon##, we can state this as
\frac{\mathrm d}{\mathrm d \epsilon}\bigg|_{\epsilon=0} S[q+\epsilon\eta] =0 \text{ .}
Now we can insert the definition of ##S[q]## and (assuming everything behaves nicely) move the derivative under the integral:
\frac{\mathrm d}{\mathrm d \epsilon}\bigg|_{\epsilon=0} S[q+\epsilon\eta] = \int_{t_a}^{t_b} \frac{\partial}{\partial\epsilon}\bigg|_{\epsilon=0} L(q(t)+\epsilon\eta(t),\dot q(t)+\epsilon\dot\eta(t))\mathrm d t
Note that the derivative acts on a function of the form ##f(g(\epsilon,t),h(\epsilon,t))##, so we can just apply the chain rule:
\frac{\partial}{\partial\epsilon}\bigg|_{\epsilon=0} L(q(t)+\epsilon\eta(t),\dot q(t)+\epsilon\dot\eta(t)) \\= \left[\frac{\partial L(q(t)+\epsilon\eta(t),\dot q(t)+\epsilon\dot\eta(t))}{\partial (q(t)+\epsilon\eta(t))}\frac{\partial (q(t)+\epsilon\eta(t))}{\partial\epsilon}+\frac{\partial L(q(t)+\epsilon\eta(t),\dot q(t)+\epsilon\dot\eta(t))}{\partial (\dot q(t)+\epsilon\dot\eta(t))}\frac{\partial (\dot q(t)+\epsilon\dot\eta(t))}{\partial\epsilon}\right]\bigg|_{\epsilon=0}\\=\frac{\partial L(q(t),\dot q(t))}{\partial q(t)}\eta(t)+\frac{\partial L(q(t),\dot q(t))}{\partial \dot q(t)}\dot\eta(t)
Now you just need to put this back into the integral, use the standard integration by parts trick, make sure the boundary term vanishes and derive the Euler-Lagrange equations, using the fact that it should hold for all ##\eta##. I have abused notation a little bit, but I hope it is clear how this is to be understood.