Good day to you!
One can define a derivation $\frac{d}{dx}$ on the ring of formal power series by letting $\frac{d}{dx} (\sum_{k \geq 0} a_k x^k ) = \sum_{k \geq 0} k a_k x^{k-1} $. In this case, $\frac{d}{dx}(x) = 1$.
I'm wondering, is it possible to find another derivation $()^\prime$ which also results in $(x)^\prime = 1$?
How about considering the special case of the ring of formal power series over the rational numbers?
In the case $A=\Bbb{Q}[[x]]$ no this is the only one.
$D$ is a map $A\to A$ such that $D(a+b)=D(a)+D(b), D(ab)=aD(b)+bD(a)$
$D(x)=1$
This implies that $D(c_n x^n)=c_n n x^{n-1}$ for $c_n\in \Bbb{Q}$ so $D$ is the standard derivative on $\Bbb{Q}[x]$
Let $f=\sum_{n\ge 0} c_n x^n $
Write $f=\sum_{n=0}^N c_n x^n + x^{N+1} g_N$ then $$D(f)=D(\sum_{n=0}^N c_n x^n)+x^{N+1}D(g_N)+(N+1)x^N g_N= D(\sum_{n=0}^N c_n x^n)+O(x^N)$$ which implies that $$D(f)=\lim_{N\to \infty}D(\sum_{n=0}^N c_n x^n)+O(x^N)=\sum_{n\ge 0} c_n n x^{n-1}$$