Variation of the Christoffel symbols with respect to $g^{\mu\nu}$

Question

I'm trying to find the field equations for some particular Lagrangian. In the middle I faced the term

$$\frac{\delta \Gamma_{\beta\gamma}^{\alpha}}{\delta g^{\mu\nu}} \, .$$

I know that

$$\delta \Gamma_{\beta\gamma}^{\alpha} = \frac{1}{2}g^{\sigma\alpha}(\nabla_{\beta}(\delta g_{\sigma\gamma}) + \nabla_{\gamma}(\delta g_{\sigma\beta}) - \nabla_{\sigma}(\delta g_{\beta\gamma})) \, .$$

I have two questions:

1. Is the expression for $\delta \Gamma_{\beta\gamma}^{\alpha}$ somehow related to $\frac{\delta \Gamma_{\beta\gamma}^{\alpha}}{\delta g^{\mu\nu}}$?

2. The idea at the end is to have terms like $$\frac{\delta \mathcal{L}}{\delta g^{\alpha\beta}} = 0$$ and thus make the variation of the action invariant under $\delta g^{\alpha\beta}$. So in simple words, is there is any way to have the term $\delta g^{\alpha\beta}$ put of the variation of the Christoffel symbol?

Answer 1

Possible hint with regards to your first question:

Notice that $$\frac{\delta \Gamma^a_{bc}}{\delta g^{\mu \nu}} \equiv \frac{\delta \Gamma^a_{bc}}{\delta\sqrt{-g}}\frac{\delta\sqrt{-g}}{\delta g^{\mu\nu}}.$$

It then follows that (see this answer and this wiki)

$$ \frac{\delta\sqrt{-g}}{\delta g^{\mu\nu}} = -\frac{1}{2}g_{\mu\nu}\sqrt{-g}.$$

Hopefully this can project the answer in the right direction. Or at least bring the question up on the feed again.