Why is the differential of a differential 1-form (or the wedge product) defined as it is? 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:
$$d\omega_{\mathfrak{x}}=dP_{\mathfrak{x}}dx+dQ_{\mathfrak{x}}dy$$
$$=\left(\frac{\partial P}{\partial x}dx+\frac{\partial P}{\partial y}dy\right)_{\mathfrak{x}}dx+\left(\frac{\partial Q}{\partial x}dx+\frac{\partial Q}{\partial y}dy\right)_{\mathfrak{x}}dy.$$
$$=\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)_{\mathfrak{x}}dxdy.$$
 A: Upon review, I discovered that Edwards gives his excuse for the definition of the product of differential forms being the implied wedge product.
Note on notation: Edwards represents the absolute value of the Jacobian by
$$\left|\frac{d\mathfrak{x}}{d\mathfrak{u}}\right|.$$
After deriving the change of variables forumula for an integral; given as

$$\int_{T\left[Q\right]}f\left[\mathfrak{x}\right]d\mathfrak{x}=\int_{Q}f\left[T\left[\mathfrak{u}\right]\right]\left|\frac{d\mathfrak{x}}{d\mathfrak{u}}\right|d\mathfrak{x},$$

Edwards infroms his reader:

This observation leads to an interpretation of the change of variables formula as the result of a “mechanical” substitution procedure. Suppose $T$ is a differentiable mapping from $uv$-space to $xy$-space. In the integral $\int_{T\left[Q\right]}f\left[x,y\right]dxdy$ we want to make the substitutions
$$dx=\frac{\partial x}{\partial u}du+\frac{\partial x}{\partial v}dv,dx=\frac{\partial y}{\partial u}du+\frac{\partial y}{\partial v}dv$$
suggested by the chain rule. In formally multiplying together these two “differential forms,” we agree to the conventions
$$dudu=dvdv=0\text{ and }dudv=-dvdu,$$
for no other reason than that they are necessary if we are to get the “right” answer.

Emphasis on no is in the original.  So I will emphasize the entire phrase:

for no other reason than that they are necessary if we are to get the “right” answer.

I do not like that "justification", but my reasons should be addressed in a different post.
