This is from Calculus in Manifold by Michael Spivak Chapter 4 Integration on Chains. If $f: Rn -> R$ is differentiate, then $Df(p)\in \wedge^1(R^n)$. 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 $R^3$ 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 $dx^1(p), . . . ,dx^n(p)$ is just the dual basis to $(e_1)_p, . . . ,(e_n)_p$. Thus every $k-form$ $\omega$ can be written $\omega=\sum_{i_1<...<i_k} \omega_{i_1}..._{i_k}dx^{i_1}\wedge..\wedge dx^{i_k}$
This is my first Question here.thanks in advance
As you can see, the domain of $df(p)$ is the tangent space at p (thus typical input being $v_p$) while that of $Df(p)$ is plain $\mathbb{R}^n$. This modification Spivak asks you to notice.
What is the significance of such trick? In $\mathbb{R}^n$ there is so natural an isomorphism between the tangent space at each point and $\mathbb{R}^n$ itself that $df$ and $Df$ look almost the same. But when you deal with wider class of spaces, you don't have such pleasing identification in general. Imagine how you would define a differential on a curved space like $S^n$!
To do calculus on such spaces(which is the author's goal as the title of the book suggests) you really need the notion of tangent space and must define differential forms in terms of it.