I'm used to calculating the change in the metric due to a gauge transformation in the following way:

The gauge transformation up to linear order is

\begin{equation} x^\mu \rightarrow x' ^\mu =x^\mu + \xi^\mu \end{equation}

If I think of the metric as a tensor, then the following identity holds

\begin{equation} g'_{\mu\nu}(x')=\frac{\partial x^\alpha}{\partial x'^\mu}\frac{\partial x^\beta}{\partial x'^\nu}g_{\alpha\beta}(x) \end{equation}

To linear order the coordinate change is just $\frac{\partial x^\alpha}{\partial x'^\mu}=\delta^\alpha_\mu-\xi^\alpha_{\ \ \ ,\mu}$ so we get the usual

\begin{equation} g'_{\mu\nu}(x')=g_{\mu\nu}(x)-\xi_{\mu,\nu}-\xi_{\nu,\mu} \end{equation}

Eq. 7.13 on Carroll's Spacetime and Geometry claims that the metric is corrected by $-\xi_{\mu;\nu}-\xi_{\nu;\mu}$ where the $;$ indicates a covariant derivative instead of a flat one. Since he is calculating this in the context of linearized gravity he throws the covariant derivative and ends up with the same result as I have. However, I was wondering if there's a way to get the covariant derivative with the tensorial method I'm using here. He uses a more complicated derivation involving pullbacks and Lie derivatives.