Let $\mathcal{M}$ be a Riemannian manifold and let $X:\mathcal{M} \rightarrow \mathcal{T}\mathcal{M}$ be a smooth vector field, i.e., $X(p) \in \mathcal{T}_p\mathcal{M}$ $\forall p \in \mathcal{M}$. Is it true that the covariant derivative along a curve $\gamma$ is the same thing as the differential of $X$ applied to the tangent vector along $\gamma$? In other words, does the following hold: \begin{equation} \frac{D X}{dt} |_{t=t_p} = D_p X [\gamma'], \end{equation}
where $\gamma'$ is the tangent vector along $\gamma$ at $\gamma(t_p) = p$.
Sofar I've tried to show that the differential is the Levi-Civita connection. Showing that it defines a connection, seems to work by using the chain rule. For this let $f \in C^\infty(\mathcal{M})$. Then $f(p)Y(p)$ is well defined and we have \begin{align} D_p f Y [X] &= (D_v (\cdot)Z, \, D_Z v(\cdot) )_{(f(p),Y(p))} \cdot \begin{pmatrix} D_p f(\cdot) \\ D_p Y(\cdot) \end{pmatrix} [X] \\ & = (Z, \, v)_{(f(p),Y(p))} \cdot \begin{pmatrix} D_p f(\cdot) \\ D_p Y(\cdot) \end{pmatrix} [X] = (Y(p), \, f(p)) \cdot \begin{pmatrix} D_p f(\cdot) \\ D_p Y(\cdot) \end{pmatrix} [X]\\ & = Y(p) D_p f [X] + f(p) D_p Y [X] = (Xf)_p Y(p) + f(p)D_p Y[X] \end{align} So indeed, this defines a connection.
However, I get stuck when trying to show symmetry and metric compatibility.