In his Advanced Calculus of Several Variables, Edwards defines the differential $d\omega$ of a differential 1-form $\omega=Pdx+Qdy$ to be
$$d\omega\equiv\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy.$$
But that definition seems cut from whole cloth. He gives some motivating examples before and after the definition, but they don't give me a good intuitive sense of what it really means.
He has mumbled a bit about the wedge product, but not by that name. He does not use it to derive the differential of a 1-form.
Why is $d\omega$ defined this way?
Or why is the wedge product used to multiply differential forms?
Or is there a way to picture either $d\omega$ or the wedge product? I note that, if the wedge product has already been introduced as
$$dx^{1}\wedge dx^{1}=dx^{2}\wedge dx^{2}=0,$$
$$dx^{1}\wedge dx^{2}=-dx^{2}\wedge dx^{1}=1,$$
and $$dx^{i}dx^{j}\equiv dx^{i}\wedge dx^{j}$$ has been asserted, the differential of $\omega$ can be derived as follows:
Start with the definition of differentiability
$$0=\lim_{\Delta\mathfrak{x}\to\mathfrak{0}}\frac{\Delta\omega_{\mathfrak{x}}\left[\Delta\mathfrak{x}\right]-d\omega_{\mathfrak{x}}\left[\Delta\mathfrak{x}\right]}{\left|\Delta\mathfrak{x}\right|}$$
$$=\lim_{\Delta\mathfrak{x}\to\mathfrak{0}}\frac{\Delta P_{\mathfrak{x}}\left[\Delta\mathfrak{x}\right]dx+\Delta Q_{\mathfrak{x}}\left[\Delta\mathfrak{x}\right]dy-d\omega_{\mathfrak{x}}\left[\Delta\mathfrak{x}\right]}{\left|\Delta\mathfrak{x}\right|}.$$
Take the constant objects $dx,dy$ outside the limits, but keep them in the same relative positions
$$0=\lim_{\Delta\mathfrak{x}\to\mathfrak{0}}\left[\frac{\Delta P_{\mathfrak{x}}\left[\Delta\mathfrak{x}\right]}{\left|\Delta\mathfrak{x}\right|}\right]dx+\lim_{\Delta\mathfrak{x}\to\mathfrak{0}}\left[\frac{\Delta Q_{\mathfrak{x}}\left[\Delta\mathfrak{x}\right]}{\left|\Delta\mathfrak{x}\right|}\right]dy-\lim_{\Delta\mathfrak{x}\to\mathfrak{0}}\left[\frac{d\omega_{\mathfrak{x}}\left[\Delta\mathfrak{x}\right]}{\left|\Delta\mathfrak{x}\right|}\right]$$
Apply the defintion of differentiability to P:
$$\lim_{\Delta\mathfrak{x}\to\mathfrak{0}}\left[\frac{\Delta P_{\mathfrak{x}}\left[\Delta\mathfrak{x}\right]}{\left|\Delta\mathfrak{x}\right|}-\frac{dP_{\mathfrak{x}}\left[\Delta\mathfrak{x}\right]}{\left|\Delta\mathfrak{x}\right|}\right]=0$$
$$\nabla\left[P\left[\mathfrak{x}\right]\right]\cdot\mathfrak{v}=dP_{\mathfrak{x}}\left[\mathfrak{v}\right]$$
$$=\left(\frac{\partial P}{\partial x}dx+\frac{\partial P}{\partial y}dy\right)\left[\mathfrak{v}\right]$$
$$=\frac{\partial P}{\partial x}dx\left[\mathfrak{v}\right]+\frac{\partial P}{\partial y}dy\left[\mathfrak{v}\right]$$
$$=\frac{\partial P}{\partial x}v^{x}+\frac{\partial P}{\partial y}v^{y}.$$
So
$$dP_{\mathfrak{x}}=\left(\frac{\partial P}{\partial x}dx+\frac{\partial P}{\partial y}dy\right).$$
The same applies to $dQ_{\mathfrak{x}}$.
The differential is now:
$$d\omega_{\mathfrak{x}}=dP_{\mathfrak{x}}dx+dQ_{\mathfrak{x}}dy$$
$$=\left(\frac{\partial P}{\partial x}dx+\frac{\partial P}{\partial y}dy\right)_{\mathfrak{x}}dx+\left(\frac{\partial Q}{\partial x}dx+\frac{\partial Q}{\partial y}dy\right)_{\mathfrak{x}}dy.$$
$$=\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)_{\mathfrak{x}}dxdy.$$