I'm trying to derive the equations of motion from the Nambu-Goto action for a general target space metric, and I know that the final result should have a term containing a Christoffel symbol. I couldn't find many resources online which go through this derivation for a general metric, most just assume a flat spacetime. I did, however, find this and this, both of which use a general metric. I have gotten the same expressions as the first of these. So I have the following
$$ T \partial_c(\sqrt{-h} h^{ac} G_{\alpha \mu} \partial_a X^\alpha) -\frac{T}{2}\sqrt{-h} h^{ab} (\partial_\mu G_{\alpha \beta}) \partial_a X^\alpha \partial_b X^\beta = 0$$
where $\partial_a, \partial_b, \partial_c$ are derivatives with respect to the coordinates $\sigma^i$ on the worldsheet, while $\partial_\mu$ is a derivative with respect to the spacetime coordinates $X^\mu$. I assume I can then pull the $G_{\alpha \mu}$ on the left hand side out of the parenthesis, since $G_{\alpha \mu}$ does not depend on the worldsheet coordinates. Then I multiply by $G^{\mu \lambda}$ and get
$$ \partial_c\left(\sqrt{-h} h^{ac} \partial_a X^\lambda\right) - \frac{1}{2}\sqrt{-h}h^{ab}G^{\mu \lambda}\left(\partial_\mu G_{\alpha \beta}\right) \partial_a X^\alpha \partial_b X^\beta = 0 $$
The factor $-\frac{1}{2}G^{\mu \lambda} \left(\partial_{\mu} G_{\alpha \beta}\right)$ is supposed to equal the Christoffel symbol $\Gamma^{\lambda}_{\alpha \beta}$, and while I see the resemblance I can't quite see how I'm supposed to show it. The Christoffel symbol is given by
$$ \Gamma^\lambda_{\alpha \beta} = \frac{1}{2}G^{\lambda \mu}\left(\partial_\alpha G_{\mu \beta} + \partial_\beta G_{\mu \alpha} - \partial_\mu G_{\alpha \beta}\right) $$
and so it contains the derivative that is needed, but it also has two extra terms. I don't understand how these are supposed to disappear. It's probably quite obvious, since nobody else seems to have a hard time seeing this, but I just can't see it myself.
Edit:
I've solved it now, and as I suspected it should have been painfully obvious. The assumption that I could simply pull $G_{\alpha \mu}$ out of the parenthesis was obviously not valid, since
$$ \partial_c G_{\alpha \mu} = \frac{\partial G_{\alpha \mu}}{\partial X^\nu}\partial_c X^\nu = \partial_\nu G_{\alpha \mu} \partial_c X^\nu $$
So this gives another derivative of the metric. And if we notice that the Christoffel symbol in the end will be multiplied by $\partial_a X^\alpha \partial_b X^\beta$ the first two terms in the expression for $\Gamma^\lambda_{\alpha \beta}$ are equal, by symmetry. And so the two derivatives of the metric that we have end up matching the derivatives in the Christoffel symbol.
