Proof that if $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ is a smooth function, then d$f_p = D f(p)$, under the natural identification $T_a\mathbb{R}^n \backsimeq \mathbb{R}^n$.
I was considering that taking d$f_p : T_p (\mathbb{R}^n) \rightarrow T_p(\mathbb{R}^m$, $L_n(h): \mathbb{R}^n \rightarrow T_p(\mathbb{R}^m)$, where it is the isomorphism between vector spaces such that for $h \in \mathbb{R}^n$, $L_n(h) := D(\cdot)(p)\cdot h$, I mean, $L_n(h)$ is the derivation that to each smooth function $v:U \rightarrow \mathbb{R}$ (remember that U is a neighborhood of $p$), $L_n (h)v = Dv (p) \cdot h$, (where $Dv(p)\cdot h$ is the gradient of $v$ in $p$). So, I'm proving that
$$L_m \circ Df(p) = \text{d}f_p \circ L_n$$
But I got stucked with the way that I need to define this compositions, I mean, tp proof the equality of the equation above and get what we need, thatd$f_p = D f(p)$. If someone can help me to describe more or how can I achieve it... Thanks
Elements of $T_{f(p)}\mathbb{R}^n$ can be viewed as derivations on $C^\infty(\mathbb{R}^m, \mathbb{R})$. To prove $L_m \circ Df_p=df_p\circ L_n$, let $h\in \mathbb{R}^n$ and $w\in C^\infty(\mathbb{R}^m, \mathbb{R})$.
$\begin{equation} df_p(L_n(h))(w) = L_n(h)(w\circ f) = \frac{d}{dt}\bigg\rvert_0 w(f(p+th))= Dw_{f(p)}(Df_p(h)) \end{equation}$
$\begin{equation} L_m(Df_p(h))(w)= \frac{d}{dt}\bigg\rvert_0 w(f(p)+tDf_p(h)))= Dw_{f(p)}(Df_p(h)) \end{equation}$