Generally when working with differential forms, one assumes that they are continuously differentiable, i.e. $C^r$ for some $1\le r \le \infty$. Under this hypothesis, one can define the exterior derivative in any of the usual ways. And if we regard a continuously differentiable function $f$ as a 0-form, then its exterior derivative $\mathrm{d}f$ is exactly its "total" differential in the usual sense.
However, for a function $f:\mathbb{R}^n\to \mathbb{R}$ we have a notion of differentiability which is stronger than the existence of all partial derivatives (or even all directional derivatives), yet weaker than continuous differentiability: $f$ is differentiable at $x$ if there exists a linear map $\mathrm{d}f_x$ such that
$$ f(x+h) = f(x) + \mathrm{d}f_x(h) + \epsilon \Vert h\Vert $$
where $\epsilon\to 0$ as $h\to 0$. (Here $x,h\in\mathbb{R}^n$.)
Is there an analogous definition of a not-necessarily-continuous exterior derivative of a differential $p$-form (on $\mathbb{R}^n$, say) for $p>0$?
Edit: Apparently the word "analogous" was not clear enough. What I'm hoping for is a definition like this: a $p$-form $\omega$ is differentiable if there exists a $(p+1)$-form $\mathrm{d}\omega$ such that
$$ (\text{something involving }\omega) = \mathrm{d}\omega + (\text{something going to } 0) $$
Best Answer
Let $\omega$ be a differential $k$-form on $\mathbb{R}^n$. Let $v_1,v_2,...,v_{k+1}$ be $k$ vectors in the tangent space to $\mathbb{R}^n$ at the point $p \in \mathbb{R}^n$.
Given scaling factors $h_1,h_2,...,h_{k+1}$ of the vectors $v_1,v_2,...,v_{k+1}$, we get a parallelepiped $P_h$ defined by the vectors $h_1v_1,h_2v_2,...,h_{k+1}v_{k+1}$. Let $bP_h$ be the oriented boundary of this parallelepiped.
Then the exterior derivative of $\omega$ can be defined by the formula
$$d \omega (p)(v_1,v_2,...,v_{k+1}) = \lim_{h_i \to 0} \frac{1}{h_1h_2...h_{k+1}}\int_{bP_{h}} \omega $$
Notice that this reduces to the formula for the derivative in the case $k=1$ (In this case integration of a zero form over an oriented collection of points is just signed summation).
So I propose the definition:
Let given $k+1$ vectors $v_1,v_2,...,v_{k+1}$, let $P$ be the oriented parallelepiped spanned by these vectors.
A $k$-form $\omega$ on $\mathbb{R}^n$ is said to be differentiable if there is a $k+1$-form $d\omega$ such that
$$\int_{bP} \omega = d\omega(p)(v_1,v_2,...,v_{k+1}) + \epsilon|Vol(P)|$$
where $\epsilon \to 0$ as the $v_i \to 0$
Thinking about this a bit more, what if we try to replace the integration with just evaluation?
Just to see an example play out first, consider a one form $\omega$ on $\mathbb{R}^n$. It seems reasonable to demand the following approximation (to be made precise shortly):
$d\omega(p)(v_1,v_2) \approx \omega(p)(v_1)+\omega(p+v_1)(v_2) - \omega(p+v_2)(v_1) - \omega(p)(v_2)$
This is obtained by just making a picture of the parallelogram spanned by $v_1$ and $v_2$ and taking signed sum of the boundary.
Example:
If we take a one form $\omega = fdx +gdy$ on $\mathbb{R}^2$ then we should have
$$\begin{align} d\omega(p)(v_1,v_2) &\approx \omega(p)(v_1)+\omega(p+v_1)(v_2) - \omega(p+v_2)(v_1) - \omega(p)(v_2)\\ &=f(p)dx(v_1)+g(p)dy(v_1) + f(p+v_1)dx(v_2)+g(p+v_1)dy(v_2)-f(p+v_2)dx(v_1) - g(p+v_2)dy(v_1) - f(p)dx(v_2)-g(p)dy(v_2)\\ &=[f(p+v_1)-f(p)]dx(v_2)-[f(p+v_2)-f(p)]dx(v_1)+[g(p+v_1)-g(p)]dy(v_2)-[g(p+v_2)-g(p)]dy(v_1)\\ &\approx df(p)(v_1)dx(v_2)-df(p)(v_2)dx(v_1) + dg(p)(v_1)dy(v_2)-dg(p)(v_2)dy(v_1)\\ &=(df \wedge dx)(v_1,v_2)+(dg \wedge dy)(v_1,v_2)\\ &=(df \wedge dx+dg \wedge dy)(v_1,v_2) \end{align}$$
This is, of course, the usual formula for the exterior derivative.
So I now propose the following definition:
A $k$-form $\omega$ is said to be differentiable if there is a $k+1$ form $d\omega$ such that
$$\sum_{i=1}^{k+1} (-1)^i(\omega(p+v_i)-\omega(p))(\widehat{v_i}) = d\omega(p)(v_1,v_2,...,v_{k+1}) + \epsilon \left| \textrm{Vol}(P) \right|$$,
where $\epsilon \to 0$ as $v_1,v_2,...,v_{k+1} \to 0$, $P$ is the parallelepiped based at $p$ and defined by the vectors $v_1,v_2,...,v_{k+1}$. $\widehat{v_i}$ is the ordered list $v_1,v_2,...,v_{k+1}$ with $v_i$ left out.
The expression on the LHS of the expression is "$\omega$ applied to the oriented boundary of $p$".
In the case that $\omega$ is differentiable according to the definition we have given above, we can obtain a formula for the exterior derivative which is similar in spirit to the formula for the derivative.
Namely, we have
$$ \begin{align} d\omega(p)(v_1,v_2,...,v_{k+1}) &= \sum_{i=1}^{k+1} (-1)^i(\omega(p+v_i)-\omega(p))(\widehat{v_i}) - \epsilon \left| \textrm{Vol}(P) \right| \end{align} $$
so
$$ \begin{align} d\omega(p)(hv_1,hv_2,...,hv_{k+1}) &= \sum_{i=1}^{k+1} (-1)^i(\omega(p+hv_i)-\omega(p))(h\widehat{v_i}) - \epsilon \left| \textrm{Vol}(P_h) \right|\\ h^{k+1}d\omega &= h^k \sum_{i=1}^{k+1} (-1)^i(\omega(p+hv_i)-\omega(p))(\widehat{v_i})- \epsilon h^{k+1}\left| \textrm{Vol}(P_1) \right|\\ d\omega &= \sum_{i=1}^{k+1} (-1)^i\left[\frac{\omega(p+hv_i)-\omega(p)}{h}\right](\widehat{v_i})- \epsilon \left| \textrm{Vol}(P_1) \right| \end{align} $$
Taking limits of both sides as $h \to 0$, we have
$$d\omega = \lim_{h \to 0} \sum_{i=1}^{k+1} (-1)^i\left[\frac{\omega(p+hv_i)-\omega(p)}{h}\right](\widehat{v_i})$$
In the case that $\omega$ is expressed in the form $\displaystyle\sum_{|I| = k} f_I dx_I$ (using multi index notation), you could further reduce this formula to the usual $\displaystyle\sum_{|I| = k} d(f_I) \wedge dx_I$