$F(p)=(F_1(p),...F_r(p))$ smooth map, Why is it true that $dF_p(X_p)=(X_p(F_1),...,X_p(F_r))$?

Question

Let $F:M\to \Bbb R^r$,smooth map, dim(M)=m

s.t $F(p)=(F_1(p),...F_r(p))$ so the components are $F_k:M\to \Bbb R$

Why is it true that

$dF_p(X_p)=(X_p(F_1),...,X_p(F_r))$?

I've trying to prove it without success and I also searched in Lee's smooth manifolds book, but I can't find this result.Any help is appreaciated.

Answer 1

Write $X_p=X_p^{(i)}\frac{\partial}{\partial x^i}|_p$

Following William's suggestion and using eq 3.9:

$dF_p(X_p) =dF_p(X_p^{(i)}\frac{\partial}{\partial x^i}|_p) =X_p^{(i)}dF_p(\frac{\partial}{\partial x^i}|_p) =X_p^{(i)}\frac{\partial F^j}{\partial x^i}(p)\frac{\partial}{\partial y^i}|_{F(p)} =(X_p^{(i)}\frac{\partial }{\partial x^i}_p) F^j\frac{\partial}{\partial y^j}|_{F(p)} = X_pF^j\frac{\partial}{\partial y^j}|_{F(p)}=(X_pF^1,...,X_pF^r)$