Assume a smooth one-parameter family of Riemannian metrics $g_{t}$. Write $h:=\frac {\partial}{\partial t}g$. In addition, assume that the Levi-civita connection on the Riemannian manifold $(M,g_{t})$ is given. I'm not sure if my following reasoning is correct.
In local coordinates $(x^i)$, the connection is given by: $$\nabla_{X}Y=(X^{i}Y^{j}\Gamma^{k}_{ij}+X^{i}\frac{\partial Y^{k}}{\partial x^i})\frac{\partial}{\partial x^k}.$$
Since in the same local coordinates the metric $g=g_{kp}dx^{k}dx^{p}$, then we have:
$$g(\nabla_{X}Y,Z)=(g_{kp}dx^{k}dx^{p})((X^{i}Y^{j}\Gamma^{k}_{ij}+X^{i}\frac{\partial Y^{k}}{\partial x^i})\frac{\partial}{\partial x^k},Z^{p}\frac{\partial}{\partial x^{p}})=g_{kp}(X^{i}Y^{j}\Gamma^{k}_{ij}+X^{i}\frac{\partial Y^{k}}{\partial x^i})Z^{p}$$
Since, $dx^{i}\frac{\partial}{\partial x^{j}}=\delta_{i}^{j}$. Then the derivative by $t$ is: $$\partial_{t}(g(\nabla_{X}Y,Z))=\partial_{t}g_{kp}*(X^{i}Y^{j}\Gamma^{k}_{ij}+X^{i}\frac{\partial Y^{k}}{\partial x^i})Z^{p}+g_{kp}*\partial_t((X^{i}Y^{j}\Gamma^{k}_{ij}+X^{i}\frac{\partial Y^{k}}{\partial x^i})Z^{p})=$$ $$=h(\nabla_{X}Y,Z)+g(\partial_t\nabla_{X}Y,Z).$$ Are my calculations correct, or am I just making lots of mistakes? I would appreciate for any comments/corrections.