Euler-Lagrange equations for a particle in a system of coordinate caracterised by a metric $g_{ij}$ The problem
I'm trying to show that the Euler-Lagrange equations for the Lagrangian $\mathcal{L} = \frac{1}{2}m(\dot{q} \cdot \dot{q}) - U(q,t)$ can be written as
$$
m \ddot{q}^{l} + m \Gamma^{l}_{ij} \dot{q}^i \dot{q}^j = -g^{kl} U_{,k}
$$
A couple of things to note:
I'm using the notation $U_{,k} \equiv \frac{\partial U}{\partial x^k}$
The metric $g_{ij} = g_{ij}(\vec{x})$ is symmetric (i.e. $g_{ij} = g_{ji}$) and its inverse is $g^{ji}$
The scalar product of the generalized coordinate is $\dot{q} \cdot \dot{q} = g_{ij} \dot{q}^i \dot{q}^j$
Tentative of resolution
We know that the Euler-Lagrange equations for a particle $k$ are,
$$
\frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{q}^k} - \frac{\partial \mathcal{L}}{\partial q^k} = 0
$$
Then calculating the different terms,
$$
 \frac{\partial \mathcal{L}}{\partial q^k} = \frac{1}{2} g_{ij,k} \dot{q}^i \dot{q}^j - U(q,t)_{,k}
$$
Then, it's where I'm struggling I do not know how deriving
$$
\frac{\partial}{\partial{\dot{q}^k}} g_{ij} \dot{q}^i \dot{q}^j
$$
So,
$$
\frac{\partial \mathcal{L}}{\partial \dot{q}^k} = ??
$$
 A: I've eventually found the answer.
So we have :
$$
\mathcal{L} = \frac{1}{2} m g_{ij}(\vec{x}(t))\dot{q}^i\dot{q}^j = - U(q,t)
$$
Computing the different terms in Euler-Lagrange equation for a particle $k$:
$$
\frac{\partial \mathcal{L}}{\partial \dot{q}^k} = \frac{1}{2}mg_{ij}\dot{q}^i\delta^j_k + \frac{1}{2}mg_{ij}\delta^i_k\dot{q}^j = \frac{1}{2} m g_{ik} \dot{q}^i + \frac{1}{2} g_{kj} \dot{q}^j = m g_{ik} \dot{q}^i
$$
$$
\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{q}^k} = mg_{ik,l} \dot{q}^l \dot{q}^i + mg_{ik} \ddot{q}^i 
$$
$$
\frac{\partial \mathcal{L}}{\partial q^k} = \frac{1}{2}mg_{ij,k}\dot{q}^i\dot{q}^j - U(q,t)_{,k}
$$
Putting these together to construct the Euler-Lagrange equation:
$$
mg_{ik,l} \dot{q}^l \dot{q}^i + mg_{ik} \ddot{q}^i  - \frac{1}{2}mg_{ij,k}\dot{q}^i\dot{q}^j + U(q,t)_{,k} = 0
$$
In the final form, we have $-g^{kl}U_{,k}$ so we observe that it is a good idea to multiply the whole expression by $g^{kl}$:
$$
mg_{ik,l} g^{kl} \dot{q}^l \dot{q}^i + mg_{ik} g^{kl} \ddot{q}^i  - \frac{1}{2}mg_{ij,k} g^{kl} \dot{q}^i\dot{q}^j = -g^{kl} U(q,t)_{,k}
$$
Observing that $g_{ik}g^{kl} = \delta^l_i$ and so $\delta^l_i \ddot{q}^i = \ddot{q}^l$:
$$
m \ddot{q}^l + m g^{kl} \left( g_{ik,l} \dot{q}^l \dot{q}^i  - \frac{1}{2}g_{ij,k} \dot{q}^i\dot{q}^j \right) = -g^{kl} U(q,t)_{,k}
$$
Relabelling dummy indices : $l \rightarrow j$
$$
m \ddot{q}^l + m g^{kl} \left( g_{ik,j} \dot{q}^j \dot{q}^i  - \frac{1}{2}g_{ij,k} \dot{q}^i\dot{q}^j \right) = -g^{kl} U(q,t)_{,k}
$$
$$
m \ddot{q}^l + m g^{kl} \left( g_{ik,j}  - \frac{1}{2}g_{ij,k} \right) \dot{q}^i \dot{q}^j = -g^{kl} U(q,t)_{,k}
$$
Then we observe that :
$$
g_{ik,j} \dot{q}^i \dot{q}^j = \frac{1}{2} g_{ik,j} \dot{q}^i \dot{q}^j + \frac{1}{2} g_{jk,i} \dot{q}^i \dot{q}^j
$$
Where we relabelled $i \rightarrow j$ and $j \rightarrow i$ in the second term, observing that it does not change the implicit sum.
Finally,
$$
m \ddot{q}^l + m g^{kl} \left( \frac{1}{2} g_{ik,j} + \frac{1}{2} g_{jk,i} - \frac{1}{2}g_{ij,k} \right) \dot{q}^i \dot{q}^j = -g^{kl} U(q,t)_{,k}
$$
$$
m \ddot{q}^l + \underbrace{\frac{1}{2} m g^{kl} \left( g_{ik,j} + g_{jk,i} - g_{ij,k} \right)}_{\Gamma^l_{ij}} \dot{q}^i \dot{q}^j = -g^{kl} U(q,t)_{,k}
$$
A: A tip: use the Kronecker Delta in the derivative, and remember that the x coordinates are related to the generalized coordinates q, {q^dot}. The delta will basically kill that 1/2 factor, and the Christoffel symbol arises from the corrdinate transformation (parallel transport)
