How Lagrange equations imply Newton equation Sorry for my English (it's not my mother language).
 I'm self-studying Mechanics and I'm have a little problem :
We can see that in Landau's book or in Wikipedia that when we inject the Lagrangian in Euler-Lagrange equation the term $\frac{\partial v²}{\partial q}$ vanishes. So we get $$\frac{\partial L}{\partial q}= - \frac{\partial U}{\partial q}$$ 
here more details :
We want to proof that Euler-lagrange equation imply Newton's second law :
Euler-Lagrange equation state that 

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = \frac{\partial L}{\partial q}$$

And we set that, for conservative potential fields $ L= T-V(q)= \frac{1}{2}mv² - V(q) $ 
But if we inject L in Euler-Lagrange equation we will get
$$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = m\frac{dv}{dt} - \frac{d}{dt}\frac{\partial V}{\partial \dot{q}}$$
And  
$$\frac{\partial L}{\partial q} = \frac{1}{2}m\frac{\partial v²}{\partial q} + F$$
In Landau's  book the terms $\frac{\partial V}{\partial \dot{q}}$ and $ \frac{1}{2}m\frac{\partial v²}{\partial q}$ vanishes without any explanation! Then I'm asking:
Why these terms vanish?
 A: This is quite simple. To better visualize Newton's law let's not use generalized coordinates, instead, let's use Cartesian ones. Euler-Lagrange equations hence stands:
$$
\frac{d}{dt}\frac{\partial L}{\partial\dot x} - \frac{\partial L}{\partial x} = 0
$$
Since $L = T - V$, it is clear to know that:
$$
\frac{\partial L}{\partial x} = 
\frac{\partial T}{\partial x} - \frac{\partial V}{\partial x} = 
-\frac{\partial V}{\partial x}   \quad\Longrightarrow\quad
\frac{\partial L}{\partial x} = -\frac{\partial V}{\partial x} = F
$$
Now, on the other term, the potential doesn't depends on the velocity. Hence:
$$
\frac{\partial L}{\partial \dot x} = 
\frac{\partial T}{\partial \dot x} - \frac{\partial V}{\partial \dot x} = 
\frac{\partial T}{\partial \dot x} \quad\Longrightarrow\quad
\frac{\partial L}{\partial \dot x} = 
\frac{\partial T}{\partial \dot x} = p
$$
This can be easily seen:
$$
T = \frac{1}{2}m\dot x^2 \Longrightarrow
\frac{\partial T}{\partial \dot x} = m\dot x = p
$$
Taking those results and plugging into Euler-Lagrange equations, we have:
$$
\frac{dp}{dt} - F = 0
$$
Which are, Newton's law.
