I'm puzzling over the canonical derivation of GR from the Einstein-Hilbert action; getting the derivation to gel with an explicit treatment of the functional derivative isn't working out. So the derivation (drawn here from Wikipedia, though other literature is similar) begins,
$$
I = \int{\sqrt{-g}d^4x} \left[\frac{1}{2 \kappa}R + \mathcal{L}\right]
$$
and immediately proceeds to
$$
\delta I = 0 =\int d^4x \delta g_{\mu \nu}\left[\frac{1}{2\kappa}\frac{\delta\left(\sqrt{-g}R\right)}{\delta g_{\mu \nu}}+\frac{\delta\left(\sqrt{-g}\mathcal{L}\right)}{\delta g_{\mu \nu}}\right].
$$
But the Ricci scalar depends on the first and second derivatives of the metric tensor, so why do we not have factors
$$
\delta g_{\mu \nu, \alpha}~, \qquad \delta g_{\mu \nu,\alpha \beta}~,
$$
against which we vary as well? Maybe there is some identity that in this case causes these terms to vanish, but I don't see it.
Best Answer
You absolutely do have these terms. Most people just always implicitly integrate by parts, and realistically, they hide these terms in other terms, because the algebra blows up into tons of terms very quickly. A very laborious version of this is worked out in the Classical Field Theory book in the Landau and Lifschitz series.
Alternately, you can use the Palantini form of the variation, and vary the Christoffel symbols, rather than the metric.