Exterior derivative of differential p-form $\omega$ can be defined by "(p+1)-linear part of the value of $\omega$ integrated over the boundary of infinitesimal (p+1)-parallelotope".


More specifically, $d\omega(v_1,v_2,...,v_{p+1})=\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial[tv_1,tv_2,...,tv_{p+1}]}\omega$, where $[tv_1,tv_2,...,tv_{p+1}]$ is (p+1)-parallelotope spanned by $tv_1,tv_2,...,tv_{p+1}$.

This aspect of exterior derivative is already mentioned by Petya and MathCrawler, but there is no proof why it is equal to the standard definition of exterior derivative 
. So I'll give you.


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It suffices to prove that

$$d\big(f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\big)(v_1,v_2,...,v_{p+1})\\=
\sum_{i\in \{ 1,2,3,...,n \} }\frac{f(x_1,x_2,...,x_n)}{\partial x_i} x_i\wedge x_1\wedge x_2\wedge ...\wedge x_p (v_1,v_2,...,v_{p+1})$$

is equal to $$\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial[tv_1,tv_2,...,tv_{p+1}]}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p$$

Suppose $U\subset \mathbb{R^n} $ is open, $\sigma^t:[0,t]^{p+1}\to U$ is $C^{\infty}$and, $\mathbf{t} \mapsto \big(\sigma^t_1(\mathbf{t}),\sigma^t_2(\mathbf{t}),\sigma^t_3(\mathbf{t}),...,\sigma^t_n(\mathbf{t})\big)\in U$ $f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p$ is differential p-form on $U$

In order to define $\partial\sigma^t$ with induced orientation, let $d^j_-(t_1, \dots, t_{p+1})=(t_1, \dots, t_{j-1}, 0, t_{j+1}, \dots, t_{p+1}),\\
d^j_+(t_1, \dots, t_{p-1}) =  (t_1, \dots, t_{j-1}, 1, t_{j+1}, \dots, t_{p+1}).$

Then $\partial \sigma^t = \sum_{j=1}^p (-1)^j(\sigma^t \circ d^j_- - \sigma^t \circ d^j_+)$, and it can be computed as below

$$\begin{align}
&\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial\sigma}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\\
=&\lim_{t\to0}\frac1{t^{p+1}}\sum_{j=1}^p(-1)^j\int_{(\sigma^t \circ d^j_- - \sigma^t \circ d^j_+)}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\\
=&\lim_{t\to0}\frac1{t^{p+1}}\sum_{j=1}^p(-1)^j\Big(\int_{\sigma^t\circ d^j_-}
f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p -\int_{\sigma^t\circ d^j_+}
f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\Big)\\
=&\lim_{t\to0}\frac1{t^{p+1}}\sum_{j=1}^p(-1)^j\Big(\int_{[0,t]^p}f(\sigma^t\circ d^j_-)det
\begin{vmatrix}
\frac{\sigma^t_1\circ d^j_-}{\partial t_1} & \cdots & \frac{\sigma^t_1\circ d^j_-}{\partial t_{j-1}} &\frac{\sigma^t_1\circ d^j_-}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1\circ d^j_-}{\partial t_{p+1}}\\
\vdots & \ddots & \vdots  & \vdots &\ddots&\vdots \\
\frac{\sigma^t_{k}\circ d^j_-}{\partial t_1}  &\cdots & \frac{\sigma^t_k\circ d^j_-}{\partial t_{j-1}}& \frac{\sigma^t_k\circ d^j_-}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_k\circ d^j_-}{\partial t_{p+1}}\\
\vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\
\frac{\sigma^t_{p}\circ d^j_-}{\partial t_1}  & \cdots & \frac{\sigma^t_{p}\circ d^j_-}{\partial t_{j-1}}  &\frac{\sigma^t_{p}\circ d^j_-}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}\circ d^j_-}{\partial t_{p+1}} 
\end{vmatrix}dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p
-\int_{[0,t]^p}f(\sigma^t\circ d^j_+)det
\begin{vmatrix}
\frac{\sigma^t_1\circ d^j_+}{\partial t_1} & \cdots & \frac{\sigma^t_1\circ d^j_+}{\partial t_{j-1}} &\frac{\sigma^t_1\circ d^j_+}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1\circ d^j_+}{\partial t_{p+1}}\\
\vdots & \ddots & \vdots  & \vdots &\ddots&\vdots \\
\frac{\sigma^t_{k}\circ d^j_+}{\partial t_1}  &\cdots & \frac{\sigma^t_k\circ d^j_+}{\partial t_{j-1}}& \frac{\sigma^t_1\circ d^j_+}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_k\circ d^j_+}{\partial t_{p+1}}\\
\vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\
\frac{\sigma^t_{p}\circ d^j_+}{\partial t_1}  & \cdots & \frac{\sigma^t_{p}\circ d^j_+}{\partial t_{j-1}}  &\frac{\sigma^t_{p}\circ d^j_+}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}\circ d^j_+}{\partial t_{p+1}} 
\end{vmatrix}
dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p\Big)
\end{align}$$