Let $L: C^\infty(\mathbb{R}) \to C^\infty(\mathbb{R})$ be a linear operator which satisfies:

$L(1) = 0$

$L(x) = 1$

$L(f \cdot g) = f \cdot L(g) + g \cdot L(f)$

Is $L$ necessarily the derivative? Maybe if I throw in some kind of continuity assumption on $L$? If it helps you can throw the "chain rule" into the list of properties.

I can see that $L$ must send any polynomial function to it's derivative. I want to say "just approximate any function by polynomials, and pass to a limit", but I see two complications: First $\mathbb{R}$ is not compact, so such an approximation scheme is not likely to fly. Maybe convolution with smooth cutoff functions could help me here. Even if I could rig up something I am concerned that if polynomials $p_n$ converge to $f$, I may not have $p_n'$ converging to $f'$. My Analysis skills are really not too hot so I would like some help.

I am interested in this question because it is a slight variant of a characterization given here:

Why do we teach calculus students the derivative as a limit?

I am not sure whether or not those properties characterize the derivative, and they are closely related to mine.

If these properties do not characterize the derivative operator, I would like to see another operator which satisfies these properties. Can you really write one down or do you need the axiom of choice? I feel that any counterexample would have to be very weird.