By the Leibniz rule, I expected it to be
$$\delta \Gamma^\sigma_{\mu\nu} = \frac 12 (\delta g)^{\sigma\lambda}(g_{\mu\lambda,\nu}+g_{\nu\lambda,\mu}-g_{\mu\nu,\lambda}) + \frac 12 g^{\sigma\lambda}(\partial_\nu (\delta g)_{\mu\lambda}+\partial_\mu (\delta g)_{\nu\lambda}-\partial_\lambda (\delta g)_{\mu\nu}) .\tag{1}$$
However, according to Sean Carroll,
$$\delta \Gamma^\sigma_{\mu\nu} = \frac 12 g^{\sigma\lambda}(\nabla_\nu (\delta g)_{\mu\lambda}+\nabla_\mu (\delta g)_{\nu\lambda}-\nabla_\lambda (\delta g)_{\mu\nu}).$$
In other words, the first term on the RHS of (1) is not there and partials on the second term are replaced by covariant derivatives. Why?
Best Answer
The difference between two connections is a tensor, so $\delta\Gamma$ is a tensor.
Evaluate your variational formula in Riemannian normal coordinates at some arbitrary point $x_0$. Since the metric derivatives are zero at that point one gets $$ \delta\Gamma^\sigma_{\mu\nu}=\frac{1}{2}\eta^{\sigma\lambda}(\partial_\nu\delta g_{\mu\lambda}+\partial_\mu\delta g_{\nu\lambda}-\partial_\lambda\delta g_{\mu\nu}), $$ where all functions are evaluated at $x_0$
In Riemannian normal coordinates, $\partial=\nabla$, so we can rewrite $$ \delta\Gamma^\sigma_{\mu\nu}=\frac{1}{2}g^{\sigma\lambda}(\nabla_\nu\delta g_{\mu\lambda}+\nabla_\mu\delta g_{\nu\lambda}-\nabla_\lambda\delta g_{\mu\nu}). $$
This equation however is tensorial, so it must be valid at $x_0$ in other coordinates too, not just Riemannian normal coordinates.
Since $x_0$ was arbitrary, this relation must then hold for any point.