The multivector derivative
$\partial _X \equiv e^I \partial_I \equiv \partial_{\langle X \rangle_0} + e^1\partial_{X_1} + e^2 \partial_{X_2}+\dots + e^{1,2,3} \partial_{X_{1,2,3}}$
with respect to some multivector variable is defined as the linear combination of all of the basis k-vectors with the components being the partial derivative operators with respect to the variable’s components.
My question is if the operator would give the same results if you changed the coordinates, or in other words, if the following statement is true:
$\partial_{X}\equiv e^I \partial_I \equiv \partial_{\langle X \rangle_0} + e^1\partial_{X_1} + e^2 \partial_{X_2}+\dots + e^{1,2,3} \partial_{X_{1,2,3}} = \partial_{\langle X \rangle_0}+ e’^1\partial_{X’_1} + e’^2 \partial_{X’_2}+\dots + e’^{1,2,3} \partial_{X’_{1,2,3}}$
Yes, this is indeed true.
Explanation 1
This is clearer in the other definition of the Multivector Derivative (taken from Hestenes and Sobczyk "Clifford Algebra to Geometric Calculus")
There the same $\partial_X$ is defined via an integral:
$$ \partial_X f(X) = \lim_{R\rightarrow 0} \frac{1}{R}\int dSf = \lim_{|R| \rightarrow 0} \frac{I^{-1}}{|R|} \int dSf $$ With $R$ as a neighboorhood of the 'point' X and the directed integral is taken over the boundary of $R$ with the (m-1)-vector $dS$ representing a directed volume element of $\partial R$.
This definition doesn't use any coordinates and therefore you can see that it is invariant.
Explanation 2
But of course defining it via an integral isn't as clear as defining it via partial derivatives so I'll try a handwaving explanation:
The definition via the partial derivatives is:
$$ \partial_X = \sum_J a^J (a_J \star \partial_X) $$
Every change to our basis blades affects the $\partial_X$ once covarinatly through the $a_J$ and once contravariantly through the $a^J$ and as a result we recieve an object invariant to any well-defined transformations.
When you have tranformations that don't transform a n-grade object into a different grade (which you normally don't have since in rotations, translations etc. vectors stay vectors and bivectors stay bivectors) the multivector derivative is even invariant when you view the grades of the multivector derivative as seperate objects, like $<\partial_X>,<\partial_X>_1, <\partial_X>_2,...$
Only when you just view certain parts of a grade, for example $e^2\partial_{X_2}+e^3\partial_{X_3}$ it is not invariant.