$M, N, P$ smooth manifolds and $F: M \to N$, $G: N \to P$ smooth maps. Show $d(G\circ F)_p=dG_{F(p)} \circ dF_p: T_p M \to T_{G \circ F(p)}P$.

Question

Let $M, N, P$ be smooth manifolds, let $F: M \to N$, $G: N \to P$ be smooth maps, and let $p \in M$. Show $d(G\circ F)_p=dG_{F(p)} \circ dF_p: T_p M \to T_{G \circ F(p)}P$.

By definition, if $v \in T_pM$ and $f \in C^\infty(P)$, then $[d(G\circ F)_p (v)](f)=v(f\circ G \circ F)$.

However, I am having trouble showing that $[(dG_{F(p)} \circ dF_p)(v)](f)=v(f\circ G \circ F)$.

I know that $dG_{F(p)}(dF_p(v))$ is a derivation at $G(F(p))$.

Answer 1

$d(G\circ F)_p(v)(f)=v(f\circ G\circ F)=dF_p(v)(f\circ G)=dG_{F(p)}(dF_p(v))(f)$