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.

Yeah, these force it to be ordinary differentiation. We have to show that for each fixed $x_0 \in \mathbb{R}$, the composite 

$$C^\infty(\mathbb{R}) \stackrel{L}{\to} C^\infty(\mathbb{R}) \stackrel{ev_{x_0}}{\to} \mathbb{R}$$ 

is just the derivative at $x_0$. For each $f \in C^\infty(\mathbb{R})$, there is a $C^\infty$ function $g$ such that 

$$f(x) = f(x_0) + f'(x_0)(x - x_0) + (x - x_0)^2g(x)$$ 

and so $(ev_{x_0} L)(f) = ev_{x_0}(f'(x_0) + 2(x - x_0)g(x) + (x - x_0)^2 L(g)(x))$ by the properties you listed. Of course evaluation at $x_0$ kills the last two terms and one is left with $f'(x_0)$, as desired. 
