Derivations of C(X)? or Why Must Supermanifolds be Smooth? 
What are the derivations of the algebra of continuous functions on a topological manifold?

A supermanifold is a locally ringed space (X,O) whose underlying space is a smooth manifold X, and whose sheaf of functions locally looks like a graded commutative algebra which is an exterior algebra over the sheaf of smooth functions. The wikipedia article has a detailed description. The n-lab post is a little thinner, but by pointing is out I've probably set in motion events that will alter it to be the superior article. Ahh, Heisenberg Uncertainty on the internet. 
What neither of these article tell you is why must the manifold be smooth?
Once I thought I understood the reason. Now I'm not so sure. The reason I was brought up with came from looking at some examples of maps of supermanifolds. For simplicity let X be just an ordinary manifold regarded as a (trivial) supermanifold. Let $\mathbb{R}^{0|2}$ be the supermanifold whose underlying manifold is just a point and whose ring of functions is the exterior algebra $A=\mathbb{R}[\theta_1, \theta_2]$, where $\theta_i$ is an odd generator. The algebra A is four dimensional as a real algebra. 
A map of supermanifolds $\mathbb{R}^{0|2} \to X$ is the same as a map of algebras $O(X) \to A$ where $O(X)$ is functions on X (of a to-be-determined type). Since X is a trivial supermanifold this is the same as pair of linear maps $a,b$:
$$ f \mapsto a(f) + b(f) \theta_1 \theta_2$$
It is easy to see that $a$ all by itself is an algebra map $O(X) \to \mathbb{R}$. This is the same thing as evalutation at a point $x_0 \in X$. Once I knew a reference for this, but I can't seem to find it. I think it is true as stated, that algebra homomorphisms $O(X) \to \mathbb{R}$ are in bijection with the points of the manifold $X$, where $O(X)$ is either smooth or continuous functions, without taking into account any topology. If this is false, then we should take into account the topology as well. In any event, $a(f) = f(x_0)$ is evaluation at $x_0$. What is $b$?
By using the algebra property of the above assignment we see that $b$ must satisfy:
$$b(fg) = f(x_0) b(g) + b(f) g(x_0).$$
In other words, $b$ is a derivation of $O(X)$ at the point $x_0$. 
So now the argument goes something like, varying over all the map from $\mathbb{R}^{0|2}$ to $X$ we see that the functions must have all first derivatives at all points and so must be $C^1$-functions. Using more odd manifolds $\mathbb{R}^{0|4}$, $\mathbb{R}^{0|6}$, etc. we see that they must have arbitrarily high derivatives and so must in fact be smooth.
Or do they?  
In order to really make this sort of argument hold water I think you need to know something like "there aren't any derivations of the algebra of continuous functions". You need the naive continuous version of the theory to be trivial or obviously useless. Otherwise I don't see a good motivation for why one must work smoothly, why it is the only interesting situation. So my question is: 

What exactly are the derivations (taking into account topology or not) of the algebra of continous functions on a topological manifold X? 

This is surely equivalent to knowing the local question, i.e. when $X = \mathbb{R}^n$. Why is the naive theory of continuous supermanifolds badly behaved?
 A: Let $X$ be a topological space and let $d:C(X)\to C(X)$ be a not-necessarily continuous derivation. Let $f\in C(X)$ and let $x\in X$. I want to show that $d(f)(x)=0$, and to do so it is enough to consider the case in which $f$ is real-valued and non-negative, for every element of $C(X)$ is a $\mathbb C$-linear combination of such functions. But in that case there exists a function $g\in C(X)$ such that $f=g^2$, and then $d(f)(x)=2g(x)d(g)(x)=0$ because $g(x)=0$.
(Note that by [Sakai, Shôichirô. On a conjecture of Kaplansky. Tôhoku Math. J. (2)  12  1960 31--33. MR0112055 (22 #2913)] a derivation of a $C^*$-algebra is automatically continuous, so imposing continuity in this context does not change much)
A: Here's a proof from definition. (I don't think it has anything to do with compactness.) Let us show that derivation 
$\delta:C(X)\to C(X)$ vanishes for any topological manifold $X$. Indeed, $\delta(1)=0$ as usual, so it suffices to show that whenever $f\in C(X)$ vanishes at $x\in X$, so does
$\delta(f)$. For this, it is enough to write $f$ as a product $f=g_1g_2$ with $g_1(x)=g_2(x)=0$. Take $g_1=\sqrt{|f|}$, $g_2=f/g_1$, and define $g_2$ to be zero where it is undefined.
A: Assuming that the manifold has a smooth structure (not guaranteed, I know) then can't you argue as follows?  Any derivation of the continuous functions must restrict to a derivation of the subalgebra of smooth functions and we know what all of those are (i.e. differentiation), and that they don't extend to the algebra of continuous functions.  Hmm, to complete the argument as simply as possible then I want to use the fact that smooth functions are dense in continuous ones so I guess I also want to work with continuous derivations.
But, of course, I can remove my first assumption since any manifold locally has a smooth structure and derivations are local phenomena!  And I also note that Mariano quotes a result that says any derivation must be continuous.  Voila!
