Let's $X$ is a fixed smooth vector field on semi-Riemannian manifold $(M,g)$. For a symmetric 2-tensor field $s$, and for sufficiently small values of $t$, $\tilde{g}=g+ts$ is a semi_Riemannian metric on $M$. I want to compute derivative of $div(X)=tr_{\tilde{g}}(\nabla X)$ at $t=0$.
how can i compute derivative of $div(X)$ at $t=0$?
We have $$\DeclareMathOperator\divergence{div} \divergence X=X^a_{\hphantom{a};a}=X^a_{\hphantom{a},a} + X^b\Gamma_{ab}^a.$$ Differentiating with respect to $t$, we get $$\dot{\divergence X}=X^b\dot{\Gamma_{ab}^a}.$$ Since $$\Gamma_{ab}^c=\frac12 g^{cd}(g_{ad,b}+g_{bd,a}-g_{ab,d}),$$ we get $$\dot{\Gamma_{ab}^c}=\frac12 \dot{g^{cd}}(g_{ad,b}+g_{bd,a}-g_{ab,d}) + \frac12 g^{cd}(\dot{g_{ad,b}}+\dot{g_{bd,a}}-\dot{g_{ab,d}}).$$ In normal coordinates at a point $p\in M$, the first derivatives of the metric are zero and $g^{cd}=\delta^{cd}$, so that at $t=0$ $$\dot{\Gamma_{ab}^c}=\frac12 \delta^{cd}(s_{ad,b}+s_{bd,a}-s_{ab,d})$$ and hence $$\dot{\Gamma_{ab}^a}=\frac12 \sum_a(s_{aa,b}+s_{ba,a}-s_{ab,a}) = \frac12\sum_a s_{aa,b}.$$ Thus in normal coordinates at $p$ at $t=0$, $$\dot{\divergence X}=\frac12\sum_a X^b s_{aa,b}.$$ In coordinate-free notation, the result is $$\frac{d}{dt}\bigg|_{t=0}\divergence_{\tilde{g}} X = \frac12 X(\operatorname{tr}_g s).$$