Use of functional derivative in derivation of Einstein Field Equations I'm trying to understand the derivation of the Einstein field equations through the stationary action principle. I've found a pretty clear explanation on the Wikipedia article for the Einstein-Hilbert action (https://en.wikipedia.org/wiki/Einstein%E2%80%93Hilbert_action), which uses calculations based on those found in "Spacetime and Geometry: An introduction to General Relativity" by Sean M. Carroll.
I'm having trouble, though, with the notation used in the article: is the $\delta$-operator used in the article the functional derivative operator?
For example, I am interpreting the variation $\delta S$ of the action $S(g) = \int [\sqrt{-g}R + \sqrt{-g}L_m] ~ d^4x $ as
$ \lim_{\epsilon \to 0} \frac{d(S_{\epsilon})}{d\epsilon} $, where $ \epsilon \mapsto \sigma^{\epsilon}$ is a curve in the space of the metrics on spacetime, based in $g$, and $ S_{\epsilon}$ is $S(\sigma^{\epsilon})$
Then, I have interpreted terms like $\frac{\delta \sqrt{-g}}{\delta g^{\mu \nu}}$ as defined through the relation $ \delta \sqrt{-g} = \int \frac{\delta\sqrt{-g}}{\delta g^{\mu \nu}}\delta g^{\mu \nu} $ (I have found this definition on the Wikipedia article for functional derivatives, https://en.wikipedia.org/wiki/Functional_derivative).
But when we calculate $ \delta R = R_{\mu \nu} \delta g^{\mu \nu}$ and the article follows with $\frac{\delta R}{\delta g^{\mu \nu}} = R_{\mu \nu} $.
I am having problems understanding this step: why are we just "dividing" both sides of the equations for $\delta g^{\mu \nu}$ and forgetting about the integral? Am I wrong in understanding the meaning of the $\delta$ symbol? If so, where can I find a more mathematically sound explanation?
 A: For a functional $S[\phi]$ which takes in a function $\phi(x)$ as input and spits out a number, the functional derivative is defined (at least in physics circles) as
$$
\frac{\delta S}{\delta \phi(x)} = \lim_{\varepsilon \to 0} \frac{S[\phi + \varepsilon \delta_x] - S[\phi]}{\varepsilon} = \frac{d}{d\varepsilon} S[\phi + \varepsilon \delta_x] \Big|_{\varepsilon = 0}
$$
where $\delta_x(y) = \delta(x-y)$ is the Dirac delta function. More generally, if you instead consider varying $S[\phi]$ by replacing $\phi(x) + \varepsilon \eta(x)$, the derivative of $S[\phi + \varepsilon \eta]$ with respect to $\eta$ is
$$
\frac{d}{d\varepsilon} S[\phi + \varepsilon \eta] \Big|_{\varepsilon = 0} = \int dx \frac{\delta S}{\delta \phi(x)} \eta(x)
$$
You can check by replacing $\eta$ with $\delta_x$ that you recover the previous definition. As a final piece of notation, people often write $\varepsilon \eta(x) \equiv \delta \phi(x)$. Then, the first-order (in $\delta \phi$) difference in $S$ upon varying $\phi \to \phi + \delta \phi$ is defined as
$$
\delta S = S[\phi + \delta \phi] - S[\phi] = \int dx \frac{\delta S}{\delta \phi(x)} \delta \phi(x)
$$
To compare with your vector calculus intuition, you should think of $\phi$ as a vector in some extremely high-dimensional space, with the components of $\phi$ labeled by the input $x$. The first equation is then analogous to a partial derivative, the second equation is analogous to a directional derivative, and the final equation is analogous to the total differential.
