Hilbert action's invariance under general coordinate changes In an article, when considering invariance of the Hilbert action under a general coordinate change this formula appears for how the metric changes
$$\delta{}g_{\mu\nu}=\partial_{\mu}\xi^{\rho}g_{\rho\nu}+\partial_{\nu}\xi^{\rho}g_{\rho\mu}+\xi^{\rho}\partial_{\rho}g_{\mu\nu}.$$
I have (maybe naively) tried using the tensor coordinate change formula considering $x^{\alpha}~\to~ x'^{\alpha}=x^{\alpha}+\xi^{\alpha}$ (formula I guessed,  it isn't specified in the article I am reading) but I don't get there.
So, how does this formula come up?
 A: You're on the right track.
Hints:


*

*Recall that under a diffeomorphism $f$, the tensor transformation law tells us that the metric transforms as $g\to g_f$ where
\begin{align}
(g_f)_{\mu\nu}(f(x)) = g_{\alpha\beta}(x)\partial_\mu (f^{-1})^\alpha(f(x))\partial_\nu (f^{-1})^\beta(f(x))
\end{align}
which, sending $x\to f^{-1}(x)$ can be re-written as
\begin{align}
  (g_f)_{\mu\nu}(x) = g_{\alpha\beta}(f^{-1}(x))\partial_\mu (f^{-1})^\alpha(x)\partial_\nu (f^{-1})^\beta(x). \tag{$\star$}
\end{align}

*Consider an infinitesimal diffeomorphism (physics speak for a smooth, one-parameter family of diffeomorphisms that starts at the identity)
\begin{align}
  f(x)=x - \xi(x) + O(\xi^2)
\end{align}

*Notice that
\begin{align}
  f^{-1}(x) = x+\xi(x) +O(\xi^2)
\end{align}

*Plug this into the right hand side of $(\star)$ and Taylor expand about $\xi=0$ to first order.

*Recall that $\delta g = g_f - g + O(\xi^2)$, and compare to the expression you wrote down.
Addendum. (2 April 2014) Notice that the first transformation law I wrote down is a more mathematically explicit version of
\begin{align}
  (g')_{\mu\nu}(x') = g_{\alpha\beta}(x) \frac{\partial x^\alpha}{\partial x'^\mu}(x')\frac{\partial x^\beta}{\partial x'^\nu}(x')
\end{align}
since if we write $x' = f(x)$ then $x = f^{-1}(x')$ so in particular
\begin{align}
  \frac{\partial x^\alpha}{\partial x'^\mu}(x') = \frac{\partial(f^{-1})^\alpha}{\partial x'^\mu}(f(x)) = \partial_\mu(f^{-1})^\alpha(f(x))
\end{align}
where in the last equality, I have simply suppressed the prime in the derivative notation; a derivative $\partial_0$ for example simply means "take the derivative with respect to the $0^\mathrm{th}$ argument of the function."  While we usually label the zeroth argument of $f^{-1}$ with the letter $x'^0$ because we are thinking of $f^{-1}$ as the transformation that maps us from the "primed" coordinates to the "unprimed" coordinates, but this is just a dummy label, and we don't strictly need it as long as the derivative tells us which argument of the function we are differentiating with respect to.
A: Write $$g^\mathfrak{ab} = x^\mathfrak{a}_\mu x^\mathfrak{b}_\nu g^{\mu\nu}.$$
Consider everything as a function of ${x'} = x + \xi$ and Taylor expand to first order. 
