These are the definitions
Let $U$ an open subset of $\mathbb{R}^n$ and $\omega:U\to\Lambda^kT^*\mathbb{R}^n$ a smooth k-form such that in local coordinates is written as
$\omega(\underline{x})=\Sigma_{_I^{\to}}\omega_{_I^{\to}}(\underline x)dx_{_I^{\to}}$, with $\omega_{_I^{\to}}:U\to\mathbb{R}$ smooth; we define de exterior derivative of $\omega$ by
$d\omega_{\underline x}(v_1,...,v_{k+1})=\sum_{i=1}^{k+1}(-1)^{i-1}\omega_{*,\underline x}(v_i)(v_1,...,\hat v_i,...,v_{k+1})$
$\hspace{3.5cm}=\sum_{i=1}^{k+1}(-1)^{i-1}\frac{\partial\omega_{_I^{\to}}}{\partial v_i}\vert_{\underline x} dx_{_I^{\to}}(v_1,...,\hat v_i,...,v_{k+1}).$
I have to prove this property of the differential:
If $\omega=fdx_{_I^{\to}}\Rightarrow d\omega=df\wedge dx_{_I^{\to}}$ I do not know how to prove that... ($f$ is a 0-form)
I only know that
$d\omega_{\underline x}(v_1,...,v_k)=\sum_{i=1}^{k+1}(-1)^{i-1}\frac{\partial f}{\partial v_i}dx_{_I^{\to}}(v_1,...,\hat v_i,...,v_{k+1})$
$=\sum_{i=1}^{k+1}(-1)^{i-1}d_{\underline x}f(v_i)dx_{_I^{\to}}(v_1,...,\hat v_i,...,v_{k+1})$ and now?
