$(M,g)$ is 2-dimensional Riemannian manifold. $C(t):(0,a)\rightarrow M$ is a curve. $V(t)$ is a vector field along $C(t)$. Assume $$ h(t)=\exp_{c(t)} V(t) \tag{1} $$ Then, I want to know how to calculate the tangent vector of $h(t)$.
What I try: In fact, it is to calculate $\partial_t h(t)$. $$ \begin{align} \partial_t h(t) = \partial_t[\exp_{c(t)} V(t)] = \partial_t[\gamma(1,C(t),V(t))] \end{align} \tag{2} $$ where $\gamma(s,C(t),V(t))$ is geodesic, which start at $C(t)$ in the direct $V(t)$, $s$ is the parameter. For fixed $t\in(0,a)$, let $$ J(s)=\gamma(s,C(t),V(t)) \tag{3} $$ denote the parallel transport along $J(s)$ as $J^*_s: T_{J(0)}M\rightarrow T_{J(s)}M$ . Then, I guess $$ \partial_t h(t)=J_1^* \left[\frac{d}{dt}C(t)+\frac{D}{dt}V(t)\right] \tag{4} $$ where $\frac{D}{dt}$ is covariant derivative along $C(t)$. But I don't know how to show it. If it is in some book, just tell me the book is enough, thanks.