Let $G$ be a Lie group, $M$ be a manifold, $\varphi:G\times M \rightarrow M$ smooth action of $G$ over $M$ where $\varphi(g,m)=g\cdot m$. For each $\xi \in \mathfrak{g}=T_e G$ we difine the fundamental vector field $X_\xi \in \mathfrak{X}(M)$ as follows: $$X_\xi (m)=\left.\dfrac{d}{d t}\right|_{t=0}\varphi(\exp(t\xi ),m )$$ where $\exp$ is exponential map. For each $g \in \mathfrak{g}=T_e G$ we difine $\varphi_g : M \rightarrow M$ where $\varphi_g (m)=\varphi(g,m)$, note that $\varphi_g $ is smooth.
Let $\xi,\eta \in \mathfrak{g}$. Show that $$\left.\dfrac{d}{d s}\right|_{s=0}X_\eta (m)(f\circ \varphi_{\exp(s\xi )})=(X_\eta f)_{*,m}(X_\xi (m))$$ where \begin{array}{rcl} X_k f:M &\rightarrow & \mathbb{R} \\ m &\rightarrow & X_k (m)(f) \end{array} for $k=\eta, \xi$ and $(X_\eta f)_{*,m}$ is differential at $m$ of $X_\eta f$.