How are the Euler-Lagrange Equations any more coordinate-invariant than Newton's? In my experience it is often said that the Lagrangian formulation of mechanics can be much much more convenient because the form of the (E-L) equations remains the same whatever coordinates we choose, allowing us to pick as convenience dictates.  Often this is exemplified by an example featuring some system constrained to a circle or a sphere, or something. 
It is true that the Second Law $\mathbf{F} = m\ddot{\mathbf{x}}$ requires us to express the system in special "inertial" coordinates, but this is fixed easily by introducing a connection and writing $\mathbf{F} = \nabla_{\dot{\mathbf{x}}}\dot{\mathbf{x}}$.  Of course, introducing a connection is conceptually identical with declaring certain curves to be geodesics which is conceptually identical with the original speak about "inertial frames." However, it is coordinate-invariant, and as far as I'm concerned coordinate-invariant is coordinate-invariant is coordinate-invariant.
 A: Well, both work in any coordinate system, but the form of Newton's law can and does change, while the form of the Euler-Lagrange equations do not change. For a very basic example, consider Newton's second law vs the Euler-Lagrange equations in polar coordinates. Newton's second law changes significantly, and becomes
$$
\begin{align}
F_r&=m\left(\ddot{r}-r\dot{\phi}\right)\\
F_\phi &= m\left(r\ddot{\phi}+2\dot{r}\dot{\phi}\right)
\end{align}
$$
This is far different then the simple Cartesian form:
$$
\begin{align}
F_x &= m\ddot{x}\\
F_y &= m\ddot{y}
\end{align}
$$
Meanwhile, the Euler-Lagrange equations stay precisely the same in form:
$$
\begin{align}
\frac{\partial L}{\partial r}&=\frac{d}{dt}\frac{\partial L}{\partial \dot{r}}\\
\frac{\partial L}{\partial \phi}&=\frac{d}{dt}\frac{\partial L}{\partial \dot{\phi}}
\end{align}
$$
compared to 
$$
\begin{align}
\frac{\partial L}{\partial x}&=\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}\\
\frac{\partial L}{\partial y}&=\frac{d}{dt}\frac{\partial L}{\partial \dot{y}}
\end{align}
$$
So, you'll get the same results no matter what (both are equivalent), but the Euler-Lagrange equations are a whole lot easier to use then Newton's law in certain coordinate systems.
