Check that F*(Xp) is a derivation at F(p) To show linearity is simple but I am stuck on derivation



 A: So you have to show that $F_{\ast}(X_p) : C^{\infty}_{F(p)}(N) \rightarrow \mathbb{R}$ is a linear map verifying a Leibniz rule, i.e. for all $f,g \in C^{\infty}_{F(p)}(N)$ we have
$$
F_{\ast}(X_p)(fg) = g(p)(F_{\ast}(X_p)f) + f(p)(F_{\ast}(X_p)g)
$$
So let $f, g \in C_{F(p)}^\infty(N)$, et $\lambda \in \mathbb{R}$. We have 
\begin{eqnarray*}
F_*(X_p) (\lambda g + f) & = & X_p \left ( (\lambda g + f) \circ F \right ) \\
& = & X_p \left (\lambda g \circ F + f \circ F \right ) \\
& = & \lambda X_p \left (g \circ F \right ) + X_p \left ( f \circ F \right ) \\
& = & \lambda F_* X_p (g) + F_* X_p(f)
\end{eqnarray*}
So $F_*(X_p)$ is $\mathbb{R}$-linear. For Leibniz rule
\begin{eqnarray*}
F_*(X_p) (gf) & = & X_p \left ( (gf) \circ F \right ) \\
& = & X_p \left ( (g \circ F) (f \circ F )\right ) \\
& = & f \circ F(p) X_p \left ( f \circ F \right ) + g \circ F(p) X_p \left ( g \circ F \right ) \\
& = & f\left ( F(p) \right )X_p \left ( f \circ F \right ) + g\left ( F(p) \right ) X_p \left ( g \circ F \right ) \\
& = & f \left ( F(p) \right ) F_*X_p(f) + g \left ( F(p) \right ) F_* X_p(g) 
\end{eqnarray*}
which shows that $F_*(X_p)$ is a derivation at $F(p)$.
