I'm studying differential forms. In Edwards Advanced Calculus, linear functionals are defined and we learn that each linear functional is a linear combination of the dual basis $ \gamma_1, \ldots, \gamma_n$, where $$\gamma_i(v_1, \ldots, v_n) = v_i$$ ($ i=1,\ldots, n$). Then he says that he will use the standard notation $ \gamma_i = dx_i$. He also calls the $ dx_i$ differentials.
I don't understand why or how differentials are related to the dual basis (aka covectors), and I don't understand how the dual basis is related to differentiation.
Maybe my confusion partly comes from high-school calculus, where the symbol $dx$ means an infinitesimal change in $x$.
Let $U$ be an open subset of $\mathbb{R}^n$. We have coordinate functions $x_i$ which are smooth functions $U \to \mathbb{R}$ given by $x_i(a_1, \dots, a_n) = a_i$. The differential of the smooth map $x_i$ at $(a_1, \dots, a_n)$ is a linear map $dx_i : T_{(a_1, \dots, a_n)}U \to T_{a_i}\mathbb{R}$; I should really write $(dx_i)_{(a_1, \dots, a_n)}$ but I will just write $dx_i$ for the sake of brevity. As $\mathbb{R}$ is a vector space, its tangent space at any point is canonically isomorphic to itself, so $T_{a_i}\mathbb{R} \cong \mathbb{R}$. Therefore, we can regard $dx_i$ as a linear map $T_{(a_1, \dots, a_n)}U \to \mathbb{R}$, that is, as a linear functional or covector on the vector space $T_{(a_1, \dots, a_n)}U$.
If you are new to this, keep in mind that vectors and covectors only make sense in a vector space. The vector space you are working with is not $\mathbb{R}^n$ (which will later be replaced by a manifold) but a tangent space to $\mathbb{R}^n$. The above description of differentials as covectors on a tangent space works in exactly the same way for manifolds.