Let $M$ be a smooth manifold, and $\text{Fr}_{GL}(M)$ be the frame bundle. Let $\rho$ be the standard representation of $GL_{n}(\mathbb{R})$ on $\mathbb{R}^n$. I am trying to show that the associated vector bundle $E=\text{Fr}_{GL}(M)\times_\rho \mathbb{R}^n$ is isomorphic to $TM$. I have shown that the map:
$$\begin{align} F:E&\longrightarrow TM\\ [(v_1,\dots,v_n)_p,(x^1,\dots,x^n)]&\longmapsto v_ix^i \end{align}$$ where $(v_1,\cdots,v_n)$ is a frame for $T_pM$ and $(x^1,\dots,x^n)$ is some vector in $\mathbb{R}^n$. I have already shown that this map is smooth, well defined, and satisfies: $$ \begin{align} \pi_{TM}\circ F=\pi_{E} \end{align} $$ I am now trying to construct a smooth inverse. My thinking is that I can just choose any frame for $T_pM$, and then send $v\in T_pM$ to an $n$-tuple in $\mathbb{R}^n$ with that frame. In particular, each point is contained in some chart so I was thinking: $$ \begin{align} F^{-1}:TM&\longrightarrow E\\ (p,w)&\longmapsto[D_{\phi(p)}\phi^{-1},D_{p}\phi(w)] \end{align} $$ but something about this feels sketchy. I am pretty sure it is smooth, as I have already showed earlier that the map $s:U\rightarrow \text{Fr}_{GL}(M)$ given by $s(p)=D_{\phi(p)}\phi^{-1}$ is smooth, and and since the map $\text{Fr}_{GL}(M)\times \mathbb{R}^n\rightarrow E$ is smooth. I believe it also independent the choice of chart, as if $\psi$ is another chart containing $p$, then $D_{p}\phi\circ D_{\psi(p)}\psi^{-1}\in GL_{n}(\mathbb{R})$, so: $$\begin{align} [D_{\phi(p)}\phi^{-1},D_{\phi(p)}\phi(w)]=&[D_{\phi(p)}\phi^{-1}\cdot(D_{p}\phi\circ D_{\psi(p)}\psi^{-1}) ,(D_{p}\psi\circ D_{\phi(p)}\phi^{-1})\circ D_{p}\phi(w)]\\ =&[D_{\psi(p)}\psi^{-1},D_{p}\psi(w)] \end{align}$$ Is this right line of thinking or no? Any clarification or help would be appreciated.