Under a local change of coordinates $x\to x'=x+\delta x$, the metric transforms as $$g_{\mu \nu}^{\prime}\left(x^{\prime}\right)=g_{\lambda \rho}(x) \frac{\partial x^{\lambda}}{\partial x^{\prime \mu}} \frac{\partial x^{\rho}}{\partial x^{\prime \nu}}$$ I am trying to show $$\delta g_{\mu \nu}=-\frac{1}{2}\left(g_{\mu \lambda} \partial^{v} \delta x_{\lambda}+g_{\lambda \nu} \partial^{\mu} \delta x_{\lambda}+\partial^{\lambda} g_{\mu \nu} \delta x_{\lambda}\right)$$
Other references, including a few answers I found in this website, are missing this factor of one half. My own derivation also misses it. I wonder where this come from. Here is my approach:
$$g'_{\mu\nu}(x+\delta x)=g'_{\mu\nu}(x)+\delta x^\lambda\partial_\lambda g'_{\mu\nu}\approx g'_{\mu\nu}(x)+\delta x^\lambda\partial_\lambda g_{\mu\nu}$$ where in the last equation, I neglected $\delta x^\lambda\partial_\lambda\delta g_{\mu\nu}$ as it is a second order change.
Since $x'^\mu=x^\mu+\delta x^\mu$, we have $$\frac{\partial x'^\mu}{\partial x^\nu}=\delta^\mu_\nu+\frac{\partial \delta x^\mu}{\partial x^\nu}\implies \frac{\partial x^\nu}{\partial x'^\mu}=\delta^\nu_\mu-\frac{\partial \delta x^\nu}{\partial x^\mu}$$ Hence $$g'_{\mu\nu}(x)+\delta x^\lambda\partial_\lambda g_{\mu\nu}=g_{\lambda\rho}\left(\delta^\lambda_\mu-\frac{\partial \delta x^\lambda}{\partial x^\mu}\right)\left(\delta^\rho_\nu-\frac{\partial \delta x^\rho}{\partial x^\nu}\right)$$ Or equivalently $$\delta g_{\mu\nu}=g'_{\mu\nu}-g_{\mu\nu}=-\left(g_{\lambda\nu}{\partial_\mu \delta x^\lambda}+g_{\mu\lambda}{\partial_\nu \delta x^\lambda}+\partial_\lambda g_{\mu\nu}\delta x^\lambda\right)$$ which, as one can see, misses the factor of $\frac 12$.