How to calculate the $\dot\gamma$ of $\gamma(s)=\Phi_s(\Gamma(s))$?

Question

Assume $(M,g)$ is a Riemannian manifold. $\Phi_s:M\rightarrow M$ is the 1-parameter diffeomorphism group and $$ \partial_s \Phi_s = X $$ where $X$ is vector field. $\Gamma:[0,\tau]\rightarrow M$ is smooth curve. Let $$ \gamma(s) = \Phi_s(\Gamma(s)) $$ then, how to show $$ \dot\gamma = X + (\Phi_s)_* (\dot\Gamma) $$ where $(\Phi_s)_*$ is the differential of $\Phi_s$.