Why does the integral symbol disappear when applying a functional derivative?

Question

it is known that variation is defined by following:

but could anyone tell me why the integral symbol disappears after following functional derivative?

Answer 1

Define functional $$ G[g]~:=~\int \!d^4x ~\sqrt{-g(x)}.\tag{0}$$ Method 1: $$\begin{align} \int \!d^4x ~\color{red}{\frac{\delta G[g]}{\delta g_{\mu\nu}(x)}} \delta g_{\mu\nu}(x) ~=~&\delta G[g]\cr ~\stackrel{(0)}{=}~& \int \!d^4x ~\color{red}{\frac{\partial\sqrt{-g(x)}}{\partial g_{\mu\nu}(x)}} \delta g_{\mu\nu}(x).\end{align} \tag{1}$$ Method 2: $$\begin{align} \color{red}{\frac{\delta G[g]}{\delta g_{\mu\nu}(y)}} ~\stackrel{(0)}{=}~& \int \!d^4x ~\frac{\delta\sqrt{-g(x)}}{\delta g_{\mu\nu}(y)}\cr ~=~& \int \!d^4x ~\frac{\partial\sqrt{-g(x)}}{\partial g_{\kappa\lambda}(x)}\frac{\delta g_{\kappa\lambda}(x)}{\delta g_{\mu\nu}(y)} \cr ~=~& \int \!d^4x ~\frac{\partial\sqrt{-g(x)}}{\partial g_{\mu\nu}(x)}\delta^4(x\!-\!y) \cr ~=~&\color{red}{\frac{\partial\sqrt{-g(y)}}{\partial g_{\mu\nu}(y)}}.\end{align}\tag{2}$$