Angular momentum in Lagrangian Mechanics

[TheDragon]

In Newton's problem,and other central force problems in Classical Mechanics, you can get with decreasing the center of mass movement to the lagrangian:
$L=1/2m(r' ^2+r^2 \varphi'^2)-V(r)$
because $\varphi$ is cyclic, you can write:
$\frac{d}{dt}(mr^2 \varphi')=0$
or, defining the angular momentum:
$mr^2 \varphi'=l$

In order for you to understand my question we will go through two ways:

Way no.1) Now if you go ahead and use euler-lagrange equation to r you will get:
$\frac{d}{dt}(mr')=mr\varphi'^2-\frac{∂V(r)}{∂r}$
and exchanging for l you can get:
$\frac{d}{dt}(mr')=\frac{l^2}{mr^3}-\frac{∂V(r)}{∂r}$

Way no.2) But if you first excange phi and then use euler-lagrange to r you get:
$L=1/2m(r' ^2+\frac{l^2}{m^2r^2})-V(r)=1/2mr' ^2+\frac{l^2}{2mr^2}-V(r)$
and then euler lagrange for r:
$\frac{d}{dt}(mr')=\frac{∂}{∂r}(\frac{l^2}{2mr^2})-\frac{∂V(r)}{∂r}=-\frac{l^2}{mr^3}-\frac{∂V(r)}{∂r}$

The plus in way no.1 became a minus in way no.2. How can you explain it?
If you can, don't show me a solution using energy because I already know that one. I want to use only lagrangian formalism.

 

[TheDragon]

The answer to the question is that the angular momentum is a constant in relation to time, but the partial derivative in r is not zero. In other words:
$\frac{dl}{dt}=0$,

$\frac{∂l}{∂r}\neq0$.

I do think that:
$\frac{dl}{dr}=0$.