If $A = P(A) = \langle A\rangle_r$, why is $\dot \partial (\dot x \cdot A) = A\cdot \partial x$ (Geometric algebra)

Question

Let $x$ be a vector of an $n$-dimensional subspace $\mathcal A_n$ with $P$ the projection operator onto this subspace. Then, let $A= P(A) = \langle A\rangle_r$ be an $r$-vector. Then, I know that $\dot \partial (\dot x \cdot A) = A\cdot \partial x$ if $r=1$, but why is this true for general $r$?

Answer 1

It's easiest to just expand the inner product: $$ 2\partial(x\cdot A) = \partial xA - \dot\partial\hat A\dot x,\quad 2(A\cdot\partial)x = A\partial x - \dot\partial\hat A\dot x. $$ The result follows because $\partial x$ is a scalar.