How to obtain real vector from abstract tangent vector in the case of the manifold $\mathbb R^n$ I know that for every $p\in\mathbb{R}^n$ the map
\begin{align}
\Phi_p\colon\mathbb{R}^n&\to T_p\mathbb{R}^n\\
v&\mapsto D_{v,p}
\end{align}
is an isomorphism, where 
\begin{align}
D_{v,p}\colon C^{\infty}(\mathbb{R}^n,\mathbb{R})&\to\mathbb{R}\\
f&\mapsto D_{v,p}f
\end{align}
is the directional derivative.
But is there an explicit formula for ${\Phi_p}^{-1}(v)$ for a given $v\in T_p\mathbb{R}^n$?

HINT: Suppose $\Phi_p \in T_p(\Bbb R^n)$(that is, it's a derivation of the algebra $C^{\infty}(\Bbb R^n.\Bbb R)$). Now let $\alpha_i = \Phi_p(x^i)$ where $x^i$ is the coordinate projection. Now show that $\Phi_p = D_{v,p}$ where $v = (\alpha_1,\dots,\alpha_n)$.
