$df(p)$ and $Df(p)$

Question

This is from Calculus in Manifold by Michael Spivak Chapter 4 Integration on Chains. If $f:\mathbf R^n \to \mathbf R$ is differentiable, then $Df(p)\in \wedge^1(\mathbf R^n)$. By a minor modification we therefore obtain a 1-form $df$, defined by $df(p)(v_p) = Df(p)(v)$. Let us consider in particular the 1-forms $d\pi^i$. It is customary to let $x^i$ denote the function $\pi^i$ (on $\mathbf R^3$ we often denote $x^1$, $x^2$ and $x^3$ by $x$, $y$, and $z$). This standard notation has obvious disadvantages but it allows many classical results to be expressed by formulas of equally classical appearance. Since $dx^i(p)(v_p) = d\pi^i(p)(v_p) = D\pi^i(p)(v) = v^i$, we see that $dx^1(p), \ldots,dx^n(p)$ is just the dual basis to $(e_1)_p, \ldots ,(e_n)_p$. Thus every $k$-form $\omega$ can be written as $$ \omega=\sum_{i_1<\ldots<i_k} \omega_{i_1\ldots\, i_k}dx^{i_1}\wedge\ldots\wedge dx^{i_k}. $$ Spivak defines $df=\sum_{i=1}^{n}\frac{\partial f}{\partial x^i}dx^i$ and defines $Df(p)$ as derivative of $f$ at $p$. My question is what is the difference in $df(p)$ and $Df(p)$? And why will $Df(p)\in \wedge^1(\mathbf R^n)$?