What matrix corresponds to differentiation? Today, in my linear algebra class, we discussed how differentiation of polynomials of degree at most $4$ can be defined using the following matrix
$$\begin{bmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 \\
0 & 0 & 0 & 3 & 0 \\
0 & 0 & 0 & 0 & 4\end{bmatrix}$$
since differentiation is a linear transformation. My question is the following

Is there a way to represent differentiation using a matrix when we consider the space of all differentiable functions with domain and codomain $\mathbb{R}$?

Note: I've studied set theory, so if the matrix has uncountably many entries, that's fine. Axiom of choice is fine as well.
 A: The short answer is no, and the long answer includes the fact that the space of all differentiable functions on $\Bbb{R}$ is infinite dimensional. The matrix representation you give above relies on the fact that fourth degree polynomials have a basis with 5 elements: $1, x, x^2, x^3$, and $x^4$. 
A: If you have a (Hamel) basis for the space of differentiable functions, then like any linear transformation, differentiation is represented in coordinate form by a matrix.
It will be big (its dimensions would be uncountably infinite!), and unlikely to have a nice form unless you carefully construct a basis. (and there is surely not any nice construction for such a basis)
A: Speaking on a intuitive basis:
The indices of a matrix $\in \mathbb N$, so are countably infinite, and and infinite matrix can operate on  countably infinite vectors, which means functions which are expressible by a series (eg. Fourier).   
To  "extend the matrix to real indices" you shall pass to a function of two variables $K(x,y)$ so that $g(x) = \int {K(x,y)f(y)dy} $ , like for instance in Volterra's integral equations.
A: We can, for example, define the differentiation operator $D$ as a linear transformation from $C^1(\mathbb{R})$, the space of all differentiable functions on $\mathbb{R}$ with continuous derivatives, to $C^0(\mathbb{R})$, the space of all continuous functions on $\mathbb{R}$. However, the immediate obstacle is that is that both of these spaces are infinite dimensional and thus don't have a finite basis. This means there can be no representation of $D$ as a matrix.
We can however obtain such a matrix if we restrict $D$ to a finite dimensional subspace of $C^1(\mathbb{R})$. Then the range of this restriction is also necessarily finite dimensional as well. We can then obtain a matrix representation in a manner entirely analogous to the construction you saw with 4th degree polynomials. In fact, you can view the example you saw in class as a specific case of doing exactly that.
