Derivative Clifford algebra

Question

How would I take the derivative of $S=e_1x$?

Let me investigate two paths to find the derivative:

1. Taking the derivative directly:

$$ dS=e_1dx $$

2. Squaring to eliminate the basis, then taking the derivative. First, we note that $S=e_1 x\implies S/x=e_1$.

$$ S^2=(e_1 x)^2\\ S^2=x^2\\ \frac{d(S^2)}{dx}=\frac{d(x^2)}{dx}\\ 2SdS=2xdx\\ \frac{dS}{dx}=\frac{x}{S}=(e_1)^{-1}\\ dS=(e_1)^{-1}dx $$

In one case I obtain $dS=e_1dx$ and in the other, I obtain $dS=(e_1)^{-1}dx$. Does $(e_1)^{-1}=e_1$, or did I make a mistake?

Answer 1

Because $e_1^2 = 1$, $e_1^{-1} = e_1$, the end results of both methods are equivalent.