Help understanding a creative definition of the derivative based purely on integrals I am trying to understand section 3 of Here, titled "what is a derivative". At equation (3.2) Hestenes defines the derivation in relation to the integral, as follows:
$$
\partial A =\lim_{d\omega\to0} \frac{1}{d\omega} \oint_{\partial \mathcal{M}} d\sigma A \tag{1}
$$
where $d\omega$ is a volume element. Hestenes suggests that this is a very good way to think of a derivative (perhaps even the best way).
I am trying to apply this definition to the 1D case, but I am struggling to do so. Specifically, my goal is to start from (1) and obtain (2):
$$
\frac{d A[x]}{d x} = \lim_{dx\to 0} \frac{A[x+dx]-A[x]}{dx} \tag{2}
$$
My thoughts and assumptions are as follows.

*

*Since we are dealing with the 1D case, should I be using $Cl_1(\mathbb{R})$; that is, the Clifford algebra of dimension 1 over the reals with basis element $\{\hat{\mathbf{x}}_1\}$?

*Hestene claims $d\omega$ is m-vector-valued differential; that is, it is a pseudo-scalar from the tangent space of $\mathcal{M}$ evaluated at point $x \in \mathcal{M}$. In our 1D case,  $d\omega=Idx=\hat{\mathbf{x}}_1 dx$ where I is the unit pseudoscalar of $Cl_1(\mathbb{R})$

*$A$ is a function of $x$. Thus, I write $A[x]$.

*Hestenes claims that $\partial=\partial_x$ is the derivated with respect to a vector $x$. In 1D, therefore $\partial_x=\partial/\partial x$.

*Hestenes claims that $d\sigma$ is a (m-1)-valued pseudoscalar also in the tangent space of $\partial \mathcal{M}$ evaluated at point $x$. I am not sure how to downgrade $\mathcal{M}$ to $\partial \mathcal{M}$ such that it is $0$-dimensional? Am I supposed to consider $d\sigma$ as a pseudoscalar of $Cl_0(\mathbb{R})$? If so then is the answer just $d\sigma=dx$?

*Finally, Hestenes claims (starting from equation 3.2) that one needs $d\omega \wedge \partial =0$ in order to get to the geometric product. In 1D, why is $\hat{\mathbf{x}}_1dx \wedge \partial=0$? Is $\partial$ assumed in the tangent space of $\mathcal{M}$ and thus parallel to $\hat{\mathbf{x}}_1$?

*What becomes of the counter integral in 1D... does it collapses to a simple definite integral? I hope I dont have to integrate from a to b then from b to a to get back to the original point and thus to complete the "contour". If so then the integrals would simply cancel each other: $\oint_R f(x)dx = \int_a^b f(x)dx + \int_b^a f(x)dx=0$... that can't be good :(

*Since the left-most term of (1) is a derivative of A and the right-most term contains $A$ and not $A'$, then I feel the contour integral in 1D must collapse to a non-integral in order to avoid raising A to its anti-derivative.

*What is $\partial \mathcal{M}$ for a 1D manifold $\mathcal{M}$ - I am assuming it is simply an interval $[x,x+h]$, where h is an infinitesimal element?

 A: The boundary of a 1D manifold is a 0D manifold: its two endpoints. A 0-dimensional integral is just a finite sum.
I'll call the basis vector $\mathbf e_1$, and the variable position vector $\mathbf x=x\mathbf e_1$. The manifold is $M=\{x\mathbf e_1\mid a\leq x\leq b\}\cong[a,b]$, and its boundary is $\partial M=\{a\mathbf e_1,b\mathbf e_1\}\cong\{a,b\}$. Actually, $M$ and $\partial M$ should also include information about orientation. The integrals are
$$\int_Md^1\mathbf x\,A(\mathbf x)=\int_a^b\mathbf e_1dx\,A(x)=\mathbf e_1\int_a^bA(x)\,dx,$$
$$\oint_{\partial M}d^0\mathbf x\,A(\mathbf x)=({}^-1)A(a)+({}^+1)A(b)=A(b)-A(a).$$
The derivative at a point $\mathbf y$ is defined thus (with different notation; I hope it's clear):
$$\partial A(\mathbf y)=\lim_{|M|\to0\\M\to\{\mathbf y\}}\frac{1}{\left(\int_M\,d^1\mathbf x\right)}\oint_{\partial M}d^0\mathbf x\,A(\mathbf x)$$
$$=\lim_{a\to y^-\\b\to y^+}\frac{1}{(b-a)\mathbf e_1}\big(A(b)-A(a)\big)$$
$$=\mathbf e_1^{-1}\lim_{a\to y^-\\b\to y^+}\frac{A(b)-A(a)}{b-a}.$$
This is not exactly the same as the usual definition, though we could take the limit "along a certain path in interval space", such as keeping one endpoint fixed: $a=y,\,b\to y^+$.

More generally, suppose $M$ is a 1D curve in a higher-dimensional space, parametrized as $\mathbf x=f(t)$ with endpoints $\mathbf x_1=f(t_1)$ and $\mathbf x_2=f(t_2)$. The integrals are
$$\int_Md^1\mathbf x\,A(\mathbf x)=\int_{t_1}^{t_2}\frac{d\mathbf x}{dt}A(\mathbf x)\,dt,$$
$$\oint_{\partial M}d^0\mathbf x\,A(\mathbf x)=({}^-1)A(\mathbf x_1)+({}^+1)A(\mathbf x_2)=A(\mathbf x_2)-A(\mathbf x_1).$$
The derivative on $M$ is defined in terms of integrals over sub-curves $M'\subset M$ containing the given point $\mathbf y=f(t_0)\in M$:
$$\partial A(\mathbf y)=\lim_{|M'|\to0\\M'\to\{\mathbf y\}}\frac{1}{\left(\int_{M'}\,d^1\mathbf x\right)}\oint_{\partial M'}d^0\mathbf x\,A(\mathbf x)$$
$$=\lim_{t_1\to t_0^-\\t_2\to t_0^+}\frac{1}{(\mathbf x_2-\mathbf x_1)}\big(A(\mathbf x_2)-A(\mathbf x_1)\big)$$
$$=\lim_{t_1\to t_0^-\\t_2\to t_0^+}\frac{1}{\left(\frac{\mathbf x_2-\mathbf x_1}{t_2-t_1}\right)}\left(\frac{A(\mathbf x_2)-A(\mathbf x_1)}{t_2-t_1}\right)$$
$$=\frac{1}{f'(t_0)}\lim_{t_1\to t_0^-\\t_2\to t_0^+}\frac{A\big(f(t_2)\big)-A\big(f(t_1)\big)}{t_2-t_1}.$$
If $t$ is arclength, then $f'(t_0)=\frac{1}{f'(t_0)}$ is the unit tangent vector to $M$ at $\mathbf y$.
