Pincherle derivative

Remarks

The map $T \mapsto T'$ is linear and satisfies the product rule $(ST)' = S'T + ST'$, since $STx - xST = S(Tx - xT) + (Sx - xS)T$.

Example: for $D = d/dx$, $D'(f) = (xf)' - x f' = f$, so $D' = \id$. For the shift $E^h f(x) = f(x + h)$ with $h \in K$, $(E^h)'(f)(x) = (x + h) f(x + h) - x f(x + h) = h f(x + h)$, so $(E^h)' = h E^h$.

Operators with Pincherle derivative zero: a multiplication operator $M_q f = q f$ with $q \in K[x]$ commutes with $x$, so $M_q' = 0$. Conversely, $T' = 0$ forces $T(x^n) = x^n T(1)$ for all $n$, so $T = M_{T(1)}$.