The obvious (and wrong) way to define differential is

${\displaystyle d={\begin{bmatrix}dx{\dfrac {\partial }{\partial x}}\\dy{\dfrac {\partial }{\partial y}}\\dz{\dfrac {\partial }{\partial z}}\end{bmatrix}}=dx^{i}\;\partial _{i}}$

If s is a scalar function (a 0-form)

${\displaystyle {\begin{aligned}ds(x,y,z)=dx^{i}\;\partial _{i}s\quad &=\quad {\frac {\partial s(x,y,z)}{\partial x}}\,dx\quad +\quad {\frac {\partial s(x,y,z)}{\partial y}}\,dy\quad +\quad {\frac {\partial s(x,y,z)}{\partial z}}\,dz\\\\&=\quad \partial _{x}s\;dx+\partial _{y}s\;dy+\partial _{z}s\;dz\\\\&=\quad {\color {red}s_{x}}\;dx+{\color {red}s_{y}}\;dy+{\color {red}s_{z}}\;dz\\\\&=\quad (\nabla s)\cdot d\mathbf {\vec {v}} \\\\&=\quad {\text{ (a differential 1-form)}}\end{aligned}}}$

${\displaystyle d(ds)=0\quad }$ (because the curl of the gradient of a scalar function is zero)

So far so good but if s is a differential 1-form (${\displaystyle {\color {red}{\tilde {s}}_{i}}\,dx^{i}={\color {red}{\tilde {s}}_{x}}\,dx+{\color {red}{\tilde {s}}_{y}}\,dy+{\color {red}{\tilde {s}}_{z}}\,dz}$):

𝖉s${\displaystyle {\begin{aligned}={\begin{bmatrix}dx{\dfrac {\partial }{\partial x}}\\dy{\dfrac {\partial }{\partial y}}\\dz{\dfrac {\partial }{\partial z}}\end{bmatrix}}{\begin{bmatrix}{\tilde {s}}_{x}\,dx\\{\tilde {s}}_{y}\,dy\\{\tilde {s}}_{z}\,dz\end{bmatrix}}&={\begin{bmatrix}dx{\dfrac {\partial }{\partial x}}{\tilde {s}}_{x}\,dx&&dx{\dfrac {\partial }{\partial x}}{\tilde {s}}_{y}\,dy&&dx{\dfrac {\partial }{\partial x}}{\tilde {s}}_{z}\,dz\\dy{\dfrac {\partial }{\partial y}}{\tilde {s}}_{x}\,dx&&dy{\dfrac {\partial }{\partial y}}{\tilde {s}}_{y}\,dy&&dy{\dfrac {\partial }{\partial y}}{\tilde {s}}_{z}\,dz\\dz{\dfrac {\partial }{\partial z}}{\tilde {s}}_{x}\,dx&&dz{\dfrac {\partial }{\partial z}}{\tilde {s}}_{y}\,dy&&dz{\dfrac {\partial }{\partial z}}{\tilde {f}}_{z}\,dz\end{bmatrix}}&={\begin{bmatrix}0&&{\dfrac {\partial {\tilde {s}}_{y}}{\partial x}}dx\,dy&&{\dfrac {\partial {\tilde {s}}_{z}}{\partial x}}dx\,dz\\{\dfrac {\partial {\tilde {s}}_{x}}{\partial y}}dy\,dx&&0&&{\dfrac {\partial {\tilde {s}}_{z}}{\partial y}}dy\,dz\\{\dfrac {\partial {\tilde {s}}_{x}}{\partial z}}dz\,dx&&{\dfrac {\partial {\tilde {s}}_{y}}{\partial z}}dz\,dy&&0\end{bmatrix}}\end{aligned}}}$

Now 𝖉s is obviously infinitesimal, so why not call it ds? Because that would imply the existence of a scalar function f such that 𝖉s = df. If there exists a function f satisfying 𝖉s = df, then 𝖉s is called an exact form — but no such function is guaranteed to exist.

• Hence my use of pseudo-s (pseudo-slope): ${\displaystyle {\color {red}{\tilde {s}}_{x},{\tilde {s}}_{y},{\tilde {s}}_{z}}}$
• Hence also the term differential “forms” — objects having the form of a differential, but not necessarily being an actual differential.
• Hence also Franken-d 𝖉 (Unicode) and Franken-partial-d ${\displaystyle {\mathfrak {d}}}$ (\mathfrak{d}) as more appropriate.

If 𝖉s = ds (an exact form) then:

${\displaystyle {\begin{aligned}d(ds)=d^{2}s=0&={\begin{bmatrix}0&&{\dfrac {\partial \left({\frac {ds}{\partial y}}\right)}{\partial x}}dx\,dy&&{\dfrac {\partial \left({\frac {ds}{\partial z}}\right)}{\partial x}}dx\,dz\\{\dfrac {\partial \left({\frac {ds}{\partial x}}\right)}{\partial y}}dy\,dx&&0&&{\dfrac {\partial \left({\frac {ds}{\partial z}}\right)}{\partial y}}dy\,dz\\{\dfrac {\partial \left({\frac {ds}{\partial x}}\right)}{\partial z}}dz\,dx&&{\dfrac {\partial \left({\frac {ds}{\partial y}}\right)}{\partial z}}dz\,dy&&0\end{bmatrix}}\\\\&={\begin{bmatrix}0&&{\dfrac {\partial ^{2}s}{\partial x\partial y}}dx\,dy&&{\dfrac {\partial ^{2}s}{\partial x\partial z}}dx\,dz\\{\dfrac {\partial ^{2}s}{\partial y\partial x}}dy\,dx&&0&&{\dfrac {\partial ^{2}s}{\partial y\partial z}}dy\,dz\\{\dfrac {\partial ^{2}s}{\partial z\partial x}}dz\,dx&&{\dfrac {\partial ^{2}s}{\partial z\partial y}}dz\,dy&&0\end{bmatrix}}\end{aligned}}}$

Thus requiring that dx dy = - dy dx. This equation tells us that our formula for differential is wrong and it isn't difficult to guess what the correct answer is (we should be using the wedge product to limit ourselves to the anti-symmetric part of the tensor) but it doesnt really explain "why". The reason why is hiding in some other formula. See Jacobian matrix and determinant, Signed area, and Stokes' theorem.

The partial derivatives on the other hand are unaffected by order. See Symmetry of second derivatives.

${\displaystyle {\frac {\partial }{\partial x}}\left({\frac {\partial f}{\partial y}}\right)\ =\ {\frac {\partial }{\partial y}}\left({\frac {\partial f}{\partial x}}\right)\qquad {\text{or}}\qquad {\frac {\partial ^{2}\!f}{\partial x\,\partial y}}\ =\ {\frac {\partial ^{2}\!f}{\partial y\,\partial x}}}$

Another notation is:

${\displaystyle \partial _{x}\partial _{y}f=\partial _{y}\partial _{x}f\qquad {\text{or}}\qquad f_{yx}=f_{xy}}$

If we picture the first derivative as arrows then the double derivative tells us how the arrows (the 1st derivative) changes as we move perpendicular to those arrows.

• For example, ${\displaystyle {\partial ^{2}f}/{\partial x\partial y}}$ measures how the vertical arrows change as we move horizontally,
• and ${\displaystyle {\partial ^{2}f}/{\partial y\partial x}}$ measures how the horizontal arrows change as we move vertically.

Subtracting them gives the amount of rotation (within the xy plane) at that point — the curl. (See below).

It’s conceptually similar to divergence, where we instead add the horizontal and vertical compression:

• the first term measures horizontal compression,
• the second term measures vertical compression.

If a field compresses vertically but stretches horizontally then the divergence cancels out; if it compresses or expands in all directions, the divergence is nonzero.

The correct way to define Total differential:

${\displaystyle d=dx^{i}\wedge \partial _{i}={\begin{bmatrix}dx\wedge {\dfrac {\partial }{\partial x}}\\dy\wedge {\dfrac {\partial }{\partial y}}\\dz\wedge {\dfrac {\partial }{\partial z}}\end{bmatrix}}}$

If s is a differential 1-form (${\displaystyle {\tilde {s}}_{i}\,dx^{i}={\tilde {s}}_{x}\,dx+{\tilde {s}}_{y}\,dy+{\tilde {s}}_{z}\,dz}$)

𝖉s${\displaystyle {\begin{aligned}=d(f_{i})\wedge dx^{i}&=d{\tilde {s}}_{x}\wedge dx\quad +\quad d{\tilde {s}}_{y}\wedge dy\quad +\quad d{\tilde {s}}_{z}\wedge dz\\\\&={\frac {1}{2}}(\partial _{i}{\tilde {s}}_{j}-\partial _{j}{\tilde {s}}_{i})\,dx^{i}\wedge dx^{j}\\\\&=(\partial _{y}{\tilde {s}}_{z}-\partial _{z}{\tilde {s}}_{y})\,dy\wedge dz\quad +\quad (\partial _{z}{\tilde {s}}_{x}-\partial _{x}{\tilde {s}}_{z})\,dz\wedge dx\quad +\quad (\partial _{x}{\tilde {s}}_{y}-\partial _{y}{\tilde {s}}_{x})\,dx\wedge dy\end{aligned}}}$

because

${\displaystyle {\begin{aligned}d{\tilde {s}}_{x}\wedge dx\quad &=\quad (\partial _{x}{\tilde {s}}_{x}\,dx+\partial _{y}{\tilde {s}}_{x}\,dy+\partial _{z}{\tilde {s}}_{x}\,dz)\wedge dx\quad =\quad -\partial _{y}{\tilde {s}}_{x}\,dx\wedge dy-\partial _{z}{\tilde {s}}_{x}\,dx\wedge dz,\\[4pt]d{\tilde {s}}_{y}\wedge dy\quad &=\quad (\partial _{x}{\tilde {s}}_{y}\,dx+\partial _{y}{\tilde {s}}_{y}\,dy+\partial _{z}{\tilde {s}}_{y}\,dz)\wedge dy\quad =\quad \partial _{x}{\tilde {s}}_{y}\,dx\wedge dy-\partial _{z}{\tilde {s}}_{y}\,dy\wedge dz,\\[4pt]d{\tilde {s}}_{z}\wedge dz\quad &=\quad (\partial _{x}{\tilde {s}}_{z}\,dx+\partial _{y}{\tilde {s}}_{z}\,dy+\partial _{z}{\tilde {s}}_{z}\,dz)\wedge dz\quad =\quad \partial _{x}{\tilde {s}}_{z}\,dx\wedge dz+\partial _{y}{\tilde {s}}_{z}\,dy\wedge dz\end{aligned}}}$

Operation	My covector formulation (true differential-form version)		Standard vector-calculus form (metric-raised, Euclidean)
Gradient	${\displaystyle \displaystyle \nabla ^{\flat }f}$	${\displaystyle =df=(\partial _{i}f)\,e^{i}}$	${\displaystyle \displaystyle \nabla f}$	${\displaystyle =(\partial _{i}f)\,e_{i}=(df)^{\sharp }}$
Divergence	${\displaystyle \displaystyle \nabla ^{\flat }\!\cdot F}$	${\displaystyle =\delta (F^{\flat })={\star }\,d\,{\star }(F^{\flat })}$	${\displaystyle \displaystyle \nabla \!\cdot F}$	${\displaystyle ={\star }\,d\,{\star }(F^{\flat })}$
Curl	${\displaystyle \displaystyle \nabla ^{\flat }\!\times F}$	${\displaystyle ={\star }\,d(F^{\flat })}$  → (1-form)	${\displaystyle \displaystyle \nabla \times F}$	${\displaystyle ={\big (}{\star }\,d(F^{\flat }){\big )}^{\sharp }}$ → (vector)
Scalar Laplacian	${\displaystyle \displaystyle \nabla _{\flat }^{2}f}$	${\displaystyle =\delta \,df={\star }\,d\,{\star }\,df}$	${\displaystyle \displaystyle \nabla ^{2}f}$	${\displaystyle ={\star }\,d\,{\star }\,df}$
1-form Laplacian	${\displaystyle \displaystyle \Delta _{\flat }F}$	${\displaystyle =(d\delta +\delta d)(F^{\flat })=d\,{\star }\,d\,{\star }(F^{\flat })+{\star }\,d\,{\star }\,d(F^{\flat })}$
Vector Laplacian	${\displaystyle \displaystyle \nabla _{\flat }^{2}F}$	${\displaystyle ={\big (}d\,{\star }\,d\,{\star }(F^{\flat })-{\star }\,d\,{\star }\,d(F^{\flat }){\big )}}$	${\displaystyle \displaystyle \nabla ^{2}F}$	${\displaystyle =\nabla (\nabla \!\cdot F)-\nabla \times (\nabla \times F)}$

where

•  f  is a scalar field
• F is a vector field
• ${\displaystyle e_i}$: basis vector, 
• ${\displaystyle e^i }$: basis covector
• ♭ and ♯ are the musical isomorphisms
  • ${\displaystyle F^{\flat }}$: convert vector → covector
  • ${\displaystyle F^{\sharp }}$: convert covector → vector
• d: exterior derivative (antisymmetric, wedge-based)
• ${\displaystyle {\star }}$: Hodge star (maps p-forms ↔ ${\displaystyle (n\!-\!p)}$-forms)
• ${\displaystyle \delta ={\star }d{\star }}$: codifferential
• All operations above are written for flat 3-space; on a manifold, curvature terms appear in the Laplacian lines

In the Algebra of physical space (APS), also known as the Clifford algebra ${\displaystyle C\ell _{4,0}(\mathbb {R} )}$, we may represent spacetime using an orthonormal Euclidean basis ${\displaystyle {e_{0},e_{1},e_{2},e_{3}}}$ satisfying

$${\displaystyle e_{\mu }e_{\nu }+e_{\nu }e_{\mu }=2\,\delta _{\mu \nu }.}$$

Time is included on equal footing with space by defining $${\displaystyle x^{0}=ict}$$ so that all four coordinates form a Euclidean 4-vector $${\displaystyle x=x^{\mu }e_{\mu }}$$ The metric tensor is therefore the identity tensor $${\displaystyle g_{\mu \nu }=\delta _{\mu \nu }}$$

The spacetime derivative is a single geometric operator acting equally in all four directions:

$${\displaystyle \partial =e^{\mu }{\frac {\partial }{\partial x^{\mu }}}=e^{0}{\frac {1}{ic}}{\frac {\partial }{\partial t}}+e^{k}{\frac {\partial }{\partial x^{k}}}}$$

The electromagnetic field is represented by a bivector field

$${\displaystyle F=E^{i}e_{i}e_{0}-cB^{i}e_{j}e_{k}}$$

or equivalently in compact form

$${\displaystyle F=\mathbf {E} +ic\mathbf {B} }$$

where

• ${\displaystyle i=e_{1}e_{2}e_{3}}$ is the spatial pseudoscalar

The four-current is represented as a vector field

$${\displaystyle J=c\rho e_{0}+J^{k}e_{k}}$$

With these definitions, Maxwell’s four equations combine into one elegant multivector equation:

$${\displaystyle {\partial F=\mu _{0}J}}$$

Expanding this equation yields the standard differential form of Maxwell’s equations in 3-space:

$${\displaystyle {\begin{aligned}\partial F&={\big (}e_{0}\,{\tfrac {1}{c}}\partial _{t}+\nabla {\big )}\,(\mathbf {E} +i\,c\,\mathbf {B} )\\[3pt]&=e_{0}\,{\tfrac {1}{c}}\partial _{t}\mathbf {E} \;+\;e_{0}\,i\,\partial _{t}\mathbf {B} \;+\;\nabla \mathbf {E} \;+\;i\,c\,\nabla \mathbf {B} \\[3pt]&=e_{0}\!\left({\tfrac {1}{c}}\partial _{t}\mathbf {E} +i\,\partial _{t}\mathbf {B} \right)\;+\;(\nabla \cdot \mathbf {E} )\;+\;\nabla \wedge \mathbf {E} \;+\;i\,c{\big [}(\nabla \cdot \mathbf {B} )+\nabla \wedge \mathbf {B} {\big ]}\\[3pt]&=e_{0}\!\left({\tfrac {1}{c}}\partial _{t}\mathbf {E} +i\,\partial _{t}\mathbf {B} \right)\;+\;(\nabla \cdot \mathbf {E} )\;+\;i\,(\nabla \times \mathbf {E} )\;+\;i\,c\,(\nabla \cdot \mathbf {B} )\;-\;c\,(\nabla \times \mathbf {B} ).\end{aligned}}}$$

$${\displaystyle {\begin{aligned}\nabla \cdot \mathbf {E} &={\frac {\rho }{\varepsilon _{0}}},\\\nabla \times \mathbf {B} -{\frac {1}{c^{2}}}{\frac {\partial \mathbf {E} }{\partial t}}&=\mu _{0}\mathbf {J} ,\\\nabla \cdot \mathbf {B} &=0,\\\nabla \times \mathbf {E} +{\frac {\partial \mathbf {B} }{\partial t}}&=0.\end{aligned}}}$$

This form places time and space on equal footing while preserving all physical relationships, eliminating the need for an asymmetric paravector derivative. The Lorentz signature is now entirely carried by the imaginary time coordinate x^0 = i c t, allowing the metric tensor to remain the identity tensor.

Spacetime algebra

We can identify APS as a subalgebra of the spacetime algebra (STA) ${\displaystyle C\ell_{1,3}(\mathbb{R})}$, defining ${\displaystyle e_{0k}=e_{k}e_{0}}$ and ${\displaystyle I=e_{0}e_{1}e_{2}e_{3}}$.

The full derivative is now

$${\displaystyle \nabla =e^{\mu }\partial _{\mu }}$$

The field multivector is a bivector

$${\displaystyle {\begin{aligned}F&=E^{1}e_{1}e_{0}+E^{2}e_{2}e_{0}+E^{3}e_{3}e_{0}\\&\quad \quad -c(B^{1}e_{2}e_{3}+B^{2}e_{3}e_{1}+B^{3}e_{1}e_{2})\end{aligned}}}$$

and the charge and current density become a vector

$${\displaystyle {\begin{aligned}J&=J^{\mu }e_{\mu }\\&=c\rho e_{0}+J^{k}e_{k}\\&=e_{0}(c\rho -J^{k}e_{0k})\\\end{aligned}}}$$

Owing to the identity

$${\displaystyle {\begin{aligned}e_{0}\nabla &=e_{0}e^{0}\partial _{0}+e_{0}e^{k}\partial _{k}\\&=\partial _{0}+e^{0k}\partial _{k}\\&={\frac {1}{c}}{\dfrac {\partial }{\partial t}}+{\boldsymbol {\nabla }}\\\end{aligned}}}$$

Maxwell's equations reduce to the single equation

$${\displaystyle \nabla F = \mu_0 c J. }$$