$M$ is a $C^\infty$-manifold, and all functions, vector fields and differential forms on $M$ are $C^\infty$.
Let $p \in M$, and let $v \in T_pM$. Prove that $v(\phi) = d\phi_p(v)$, for all real valued functions $\phi$ on $M$.
Using Chapter 3 definition 4 from do Carmo's Differential Forms and Applications:
Let $M_1^n$ and $M_2^n$ be differentiable manifolds and let $\phi: M_1 \rightarrow M_2$ be a differentiable map. For each $p \in M$, the differential of $\phi$ at $p$ is the linear map $d\phi_p: T_pM_1 \rightarrow T_pM_2$ which associates to each $v \in T_pM_1$ the vector $d\phi_p(v) \in T_{\phi(p)}M_2$ defined as followes: Choose a differentiable curve $\alpha: (-\epsilon, \epsilon) \rightarrow M_1$, with $\alpha(0) = p,$ $\alpha'(0) = v;$ then $d\phi_p(v) = (\phi \circ \alpha)'(0).$
\begin{equation} v(\phi) = \alpha'(0)\phi = \alpha'(0)\phi + 0 = \alpha'(0)\phi + \phi'\alpha(0) = (\phi \circ \alpha)'_p(0) = \phi'v = d\phi_pv \end{equation} Can I just write my solution like this, or do I need to explain more. I'm pretty new to this notation so I have a hard time to figure out when to write the p, or when you can leave it out.