As in the title: I'm looking to take the geometric derivative with respect to $\vec{x}$ of the exponential of the commutator of $\gamma_0 \vec{k}$ and $\vec{x} \gamma^0$: $$e^{\frac{1}{2}\left[\gamma_0 \vec{k} \vec{x} \gamma^0 - \vec{x} \gamma^0 \gamma_0 \vec{k}\right]}$$ I'm working in the spacetime algebra (East Coast metric, if we need to pick a sign), so the $\gamma_\mu$ terms obey the following rules: $$ \frac{1}{2}\left[\gamma_\mu \gamma_\nu + \gamma_\nu \gamma_\mu\right] = \eta_{\mu\nu} $$ $$ \frac{1}{2}\left[\gamma^\mu \gamma_\nu + \gamma_\nu \gamma^\mu\right] = \delta^\mu_\nu $$
The reason I'm asking here is to avoid any pitfalls that may arise from misapplication of the chain rule for a noncommutative algebra. I realize that one possible approach for evaluating this derivative is to expand it as a Taylor series: $$ e^{\frac{1}{2}\left[\gamma_0 \vec{k} \vec{x} \gamma^0 - \vec{x} \gamma^0 \gamma_0 \vec{k}\right]} = \sum_{n = 0}^{\infty} \frac{\left[\gamma_0 \vec{k} \vec{x} \gamma^0 - \vec{x} \gamma^0 \gamma_0 \vec{k}\right]^n}{2^n n!} $$ $$ \begin{align} \nabla e^{\frac{1}{2}\left[\gamma_0 \vec{k} \vec{x} \gamma^0 - \vec{x} \gamma^0 \gamma_0 \vec{k}\right]} & = \sum_{n = 1}^{\infty} \frac{\left[\gamma_0 \vec{k} \vec{x} \gamma^0 - \vec{x} \gamma^0 \gamma_0 \vec{k}\right]^n}{2^n (n - 1)!} \\\\ & = \sum_{n = 0}^{\infty} \frac{\left[\gamma_0 \vec{k} \vec{x} \gamma^0 - \vec{x} \gamma^0 \gamma_0 \vec{k}\right]^{n+1}}{2^{n+1} n!} \\\\ & = \frac{1}{2}\sum_{n = 0}^{\infty} \frac{\left[\gamma_0 \vec{k} \vec{x} \gamma^0 - \vec{x} \gamma^0 \gamma_0 \vec{k}\right]^{n+1}}{2^n n!} \end{align} $$ But I don't know if I can pull the commutator expression as a term on the left, on the right, or if it's irrelevant, so I can reduce its exponent.
[edit] I'm not considering this question answered yet, but I just realized for each exponential term, it technically shouldn't matter if we pull a term out on one side or the other, as long as we do it consistently. That being said, I am wary of geometric consequences of the choice of left vs. right multiplication.
$ \renewcommand\vec\mathbf \newcommand\swedge{\mathop{\underset S\wedge}} \newcommand\slcontr{\mathop{\underset S\rfloor}} \newcommand\lcontr{\mathbin\rfloor} $I don't know which one is the "East coast" metric, though it doesn't matter that much. I also assume that $k,x$ are spacetime 4-vectors.
Throughout I will use the convention that the geometric product has less precedence than any other product, i.e. you always do geometric products last. I will also use the convention that the contraction $\lcontr$ has lower precedence than $\wedge$ and $\times$ so that e.g. $$ A\lcontr B\wedge C = A\lcontr(B\wedge C). $$
Note that $\gamma^0 = \iota\gamma_0$ where $\iota = \pm1$ depending on signature.
We can compute this in closed form. They key is understanding the relationship between the spacetime algebra (STA) and its even subalgebra which is isomorphic to the geometric algebra of 3D Euclidean space, and is the natural way of representing spatial quantities in STA. (As such, I will call it the spatial subalgebra.) We define $$ \vec x = x\wedge\gamma^0,\quad \vec k = k\wedge\gamma^0 $$ which is the spatial part of the spacetime split $$ x\gamma^0 = x\cdot\gamma^0 + \vec x $$ with $x\cdot\gamma^0$ being the time component and $\vec x$ representing a spatial vector in the spatial subalgebra.
Because scalars commute with everything, only the spatial parts matter in the commutator, yielding $$ X := \frac12[\gamma_0k, x\gamma^0] = \frac12\iota[\vec x, \vec k] = \iota\vec x\times\vec k $$ where the commutator product $\times$ is half the commutator, but overall a more meaningful operation. The quantity $\vec x\times\vec k$ is exactly the spatial wedge product of spatial vectors, yielding a spatial bivector (which is also a spacetime bivector). It immediately follows that $$ e^X = \cos|X| + X'\sin|X| $$ where $|X| = \sqrt{-X^2}$ and $X' = X/|X|$. To differentiate this, we reference Chapter 2 of Clifford Algebra to Geometric Calculus (CAGC) by David Hestenes and Garret Sobczyk. The chain rule gives us $$ \nabla e^X = \dot\nabla\dot X\cdot\partial_X e^X $$ where $\partial_X$ is the spatial bivector derivative and $A\cdot B = \langle AB\rangle_0$ is the scalar product. The overdots specify that $\dot\nabla$ is only differentiating $\dot X$. Now we get three terms, use the chain rule again on the trig functions, and apply basic identities from CAGC: $$\begin{aligned} \dot X\cdot\partial_X e^X &= {-(\dot X\cdot\partial_X|X|)}\sin|X| + \left(\dot X\cdot\partial_XX'\right)\sin|X| + X'(\dot X\cdot\partial_X|X|)\cos|X|. \\ &= \dot X\cdot X'\sin|X| + \frac1{|X|}\left(\dot X + \dot X\cdot X'X'\right)\sin|X| - X'\dot X\cdot X'\cos|X|. \end{aligned}$$ We now need to apply $\dot\nabla$ to this. We need two subexpressions $$ \dot\nabla\dot X\cdot X'\quad\text{and}\quad\dot\nabla\dot X $$
For the first, if we use $$ \swedge\quad\text{and}\quad\slcontr $$ to represent the spatial wedge product and spatial left contraction and note that the spatial subalgebra has the same scalar product as the whole spacetime algebra, then $$\begin{aligned} \dot\nabla\dot X\cdot X' &= \dot\nabla\bigl[(\dot x\wedge\gamma_0)\swedge\vec k\bigr]\cdot X' \\ &= \dot\nabla(\dot x\wedge\gamma_0)\cdot(\vec k\slcontr X') \\ &= \dot\nabla\dot x\cdot[\gamma_0\lcontr\vec k\times X'] \\ &= \gamma_0\lcontr\vec k\times X'. \end{aligned}$$ The key is that $$ \vec x\times\vec k = \vec x\swedge\vec k $$ and that, because $X'$ is a spatial bivector, the quantity $$ \vec k\slcontr X' = \vec k\times X' $$ is a spatial vector. Note that $\iota\vec x = x\wedge\gamma_0$, and that in the third and fourth lines we use the spacetime contraction.
For the second, we see $$\begin{aligned} \dot\nabla(\dot x\wedge\gamma_0)\times\vec k &= \dot\nabla\bigl[\dot x\wedge(\gamma_0\times\vec k) + (\dot x\times\vec k)\wedge\gamma_0\bigr] \\ &= 3\gamma_0\lcontr\vec k + \dot\nabla(\dot x\lcontr k\wedge\gamma^0)\wedge\gamma_0 \\ &= 3\gamma_0\vec k - \dot\nabla\dot x\cdot\gamma^0 k\wedge\gamma_0 \\ &= 3\gamma_0\vec k - \gamma_0\vec k \\ &= 2\gamma_0\vec k. \end{aligned}$$ Note that $\gamma_0\times\vec k = \gamma_0\lcontr\vec k = \gamma_0\vec k$.
All together (and replacing $X'$ with $X/|X|$) we now have $$ \nabla e^X = \frac{\sin|X|}{|X|}\gamma_0\lcontr\vec k\times X + \frac{2\sin|X|}{|X|}\gamma_0\vec k + \frac{\sin|X|-|X|\cos|X|}{|X|^3}\gamma_0\lcontr\vec k\times XX. $$ To simplify, we first start with $$\begin{aligned} \gamma_0\lcontr\vec k\times X &= \gamma_0\lcontr\bigl[\iota\vec k\slcontr\vec x\swedge\vec k\bigr] = \gamma_0\lcontr\bigl[\iota\vec k\cdot\vec x\,\vec k - \iota\vec k^2\vec x\bigr] \\ &= \iota\vec k\cdot\vec x\,\gamma_0\vec k - \iota\vec k^2\gamma_0\vec x. \end{aligned}$$ We then also multiply by $X$; since $$ \iota\vec kX = \vec k\slcontr\vec x\swedge\vec k + \vec k\swedge\vec x\swedge\vec k = \vec k\cdot\vec x\,\vec k - \vec k^2\vec x, $$ $$ \iota\vec xX = \vec x\slcontr\vec x\wedge\vec k + \vec x\swedge\vec x\swedge\vec k = \vec x^2\vec k - \vec x\cdot\vec k\,\vec x $$ we see $$\begin{aligned} \gamma_0\lcontr\vec k\times XX &= (\vec k\cdot\vec x)^2\gamma_0\vec k - \vec k\cdot\vec x\,\vec k^2\gamma_0\vec x - \vec x^2\vec k^2\gamma_0\vec k + \vec x\cdot\vec k\,\vec k^2\gamma_0\vec x \\ &= \bigl[(\vec k\cdot\vec x)^2-\vec k^2\vec x^2\bigr]\gamma_0\vec k \\ &= -|X|^2\gamma_0\vec k \end{aligned}$$
The quantities $\vec x\gamma_0 = -\gamma_0\vec x$ and $\vec k\gamma_0 = -\gamma_0\vec k$ are more natural since e.g. $\vec x\gamma_0 = x - x\cdot\gamma^0\gamma_0$ is exactly the spacetime vector representation of the spatial vector $\vec x$. Finally, we have $$ \nabla e^X = \iota S\vec k^2\vec x\gamma_0 -\Bigl[\iota S\vec k\cdot\vec x + S + \cos|X|\Bigr]\vec k\gamma_0, \tag{$*$} $$ $$ |X| = \sqrt{\vec k^2\vec x^2 - (\vec k\cdot\vec x)^2},\quad S = \frac{\sin|X|}{|X|}. $$
This has all been assuming $|X| \ne 0$. When $|X| \approx 0$ we instead see that $$ \nabla e^X \approx \nabla(1 + X) = \nabla X = 2\gamma_0\vec k. \tag{$**$} $$ For $|X| \approx 0$ to be true, either $|\vec k| \approx 0$ or there is a scalar $a$ (dependent on $x$) such that $\vec x \approx a\vec k$. When $\vec k = 0$ we see that ($*$) and ($**$) agree. When $\vec x \approx a\vec k$, we approximate ($*$) as $$ \nabla e^X \approx \iota a\vec k^3\gamma_0 - [\iota a\vec k^2 + 1]\vec k\gamma_0 = \gamma_0\vec k $$ which is inconsistent with ($**$), though perhaps merely an indication that I've made a mistake somewhere. If this is not a mistake, then that means we should restrict to $|X| > 0$.