Question about metric connection Let $\nabla$ be an affine connection on a Riemannian manifold $(S,g)$, and suppose that $\nabla$ satisfies, for all vector fields $X,Y,Z$ in $S$
\begin{equation}\label{metric_connection}
\tag{1}
Z\left\langle X,Y\right\rangle=\left\langle\nabla_{Z}X,Y\right\rangle+\left\langle X,\nabla_{Z}Y\right\rangle
\end{equation}
then we say that $\nabla $ is a metric connection with respect to $g.$
Consider a curve $\gamma:t\mapsto \gamma(t)$ on $S$ and two vector fields $X$ and $Y$ along $\gamma$. Letting $\frac{\delta X}{dt}$ and $\frac{\delta Y}{dt}$ denote the covariant derivative of $X$ and $Y$ with respect to $\nabla$ we see from $(1)$ that
\begin{equation}\label{curve_derivative}
\tag{2}
\frac{d}{dt}\left\langle X(t),Y(t)\right\rangle=\left\langle\frac{\delta X(t)}{dt},Y(t)\right\rangle+\left\langle X(t),\frac{\delta Y(t)}{dt}\right\rangle.
\end{equation}
Can someone explain how we got \eqref{curve_derivative} from \eqref{metric_connection}?
 A: Recall that we're fixing a curve $\gamma$ and $\dfrac{{\rm d}}{{\rm d} t}$ is an operator $\dfrac{{\rm d}}{{\rm d} t} : \gamma^{*}(T M) \to \gamma^{*}(T M)$ (where $\gamma^{*}(T M)$ denotes the pullback bundle of the tangent bundle of $M$ by $\gamma$ - but you can just think of it as the set of all vector fields along $\gamma$)  which satisfies the property that if $V$ is induced by a vector field $Y \in \Gamma(T M)$ (i.e. $V(t) = Y(\gamma(t))$ for all $t$ in the domain of $\gamma$), then $\dfrac{{\rm d} V}{{\rm d} t} = \nabla_{\gamma'(t)} V $ (where the latter notation makes sense because $\nabla$ is tensorial in the first entry). Also recall that vector fields act on smooth real functions defined on your manifold. That being said, we get:
$$\begin{aligned} \dfrac{{\rm d}}{{\rm d} t} \langle X(\gamma(t)), Y(\gamma(t))\rangle &= \nabla_{\gamma'(t)} \left( \langle X(\gamma(t)), Y(\gamma(t))\rangle \right) \\
&= \left\langle\frac{\delta X(t)}{dt},Y(t)\right\rangle+\left\langle X(t),\frac{\delta Y(t)}{dt}\right\rangle.\end{aligned}$$
