Gauge Variation of full metric in GR

Question

I'm going through my GR notes and came across the following.

Besides the missing integral sign in the first line I don't get the step. I thought (product rule) $$\delta_\xi (g_{\mu\nu} \dot{x^\mu}\dot{x^\nu})=\delta_\xi(g_{\mu\nu})\dot{x^\mu}\dot{x^\nu}+ 2g_{\mu\nu}\dot{x^\mu}\delta\dot{x^\nu}=\partial_\rho g_{\mu\nu}\delta\dot{x^\rho}\dot{x^\mu}\dot{x^\nu}+2g_{\mu\nu}\dot{x^\mu}\delta\dot{x^\nu}.$$

Why is there a third term ($\delta_\xi(g_{\mu\nu})\dot{x^\mu}\dot{x^\nu}$) in the first line?

Reference:GR Script, Hohm, p.24

Answer 1

The third term in eq. (183) in Hohm's notes is the (infinitesimal gauge) transformation $\delta_{\xi}g_{\mu\nu}$ of the metric field $g_{\mu\nu}$ itself. In the very next eq. (184) it is concluded that it is the Lie derivative. [All the other terms in eq. (183) are changes due to transformation $\delta_{\xi}x^{\mu}$ of the point particle position $x^{\mu}$.]