$T_p(\mathbb{R}^n)\cong D_p(\mathbb{R}^n)$
Lemma: Let $f \in C^{\infty}(U)$ where $U$ is an open subset of euclidean space, star shaped with respect to a point $p=(p^1...,p^n)$ in $U$. Then there are functions
$g_1(x)...,g_n(x)\in C^{\infty}(U)$ such that
$f(x)=f(p)+\sum_{i=1}^n(x^i-p^i)g_i(x)$ where $g_i(p)=\frac{\partial{f}}{\partial{x}^i}(p)$
My problem: All I want to show is that the map:
$\phi: T_p(\mathbb{R}^n)\rightarrow D_p(\mathbb{R}^n)$ , given by $\phi(v_p)=D_v=\sum_k v^k \frac{\partial}{\partial{x}^k}|_p.$ is surjective.
Let $D$ be a derivation at $p$. Let $[(f,V)]$ be a germ at p (this exists, take any constant function for instance). Since the restriction of a smooth map to an open domain is smooth, and $[(f,V)]=[(f,B)]$ where $B$ is some ball containing p and is contained in $V$, we may assume without loss of generality that $V$ is starshaped.
Hence, there exists $g_1,g_2...,g_i\in C^{\infty}(\mathbb{R}^n)$ such that:
$f(x)=f(p)+\sum_{i=1}^n (x^i-p^i)g_i(x)$. Applying $D$ on both sides, we get
$D[(f,V)]$ ( which we will denote by f) is equal to $D(f)=D(f(p)+ \sum_i^n (\operatorname{id}^i-p^i)g_i$) where $\operatorname{id}^i$ is the i component for the identity map.
So, since $f(p)$ is constant, and derivations are $0$ at constants, we get $D(f)=D(\sum_{i=1}^n(\operatorname{id}^i-p^i)g_i)$. So, by linearity, we get $\sum_{i=1}^nD((\operatorname{id}^i-p^i)g_i)$.
By Leibniz rule, $\sum_{i=1}^ng_i(p)D(\operatorname{id}^i-p^i)$.
How do I complete this?
Just by taking limits, we see that $g_i(p) = \frac {\partial f} {\partial x^i}\Big|_p$. (Take limit of $f = f(p) + \sum(id^i - p^i)g_i$ along $x^i$ and use the fact that $g_i$ is continuous).
Now we have $D(f)(p) = \sum v^i \frac {\partial f} {\partial x^i}\Big|_p$, where $v^i = D(id^i - p^i)$