Uniqueness of the derivative operator Let $\mathcal{D}(\mathbb{R})$ be the set of all functions $f:\mathbb{R}\to\mathbb{R}$ which are differentiable at all points, and let $\mathcal{F}(\mathbb{R})$ be the set of all functions $f:\mathbb{R}\to\mathbb{R}$.
Let's denote by $D$ the derivative operator, i.e.
$$D:\mathcal{D}(\mathbb{R})\to\mathcal{F}(\mathbb{R})$$
is such that
$$D(f)=f^{\prime}$$
It's clear that $\mathcal{F}(\mathbb{R})$ is a real vector space and that $\mathcal{D}(\mathbb{R})$ is a subspace of it. We also know that $D$ is linear and that it obeys the product rule
$$D(f\cdot g)=f\cdot D(g)+g\cdot D(f)$$
The question is: is $D$ the only such linear transformation, i.e., the only linear transformation that obeys product rule?
Of course, the $0$ operator and any real multiple of $D$, $\alpha\cdot D$ are trivial examples. But I'm interested in non-trivial examples.
 A: The differentiable functions are a subset of the continuous functions.  The polynomials are dense in the continuous functions on closed intervals.  (This is the Weierstrass approximation theorem.)  It is sufficient, therefore to consider what this operator does to polynomials.  It's action on non-polynomial differentiable functions or differentiable functions without compact support can be found by taking limits of polynomials.  (For instance, this limit could be of a sequence of polynomials,$(p_n(x))_{n \in \Bbb{N}}$, approximating a given differentiable function to within $1/n$ on the closed interval $[-n,n]$.  This is the same idea as is normally used for showing the simple functions are dense in the continuous functions.)
What happens when we put $1$ in for both functions?  $$
    D(1) = D(1 \cdot 1) = D(1) \cdot 1 + 1 \cdot D(1) = 2D(1) \text{.}
$$  This means $D(1) = 0$.  Since $D$ is linear, for any constant, $c$, $D(c) = 0$.
What happens to $x^2$ and $x^n$ (for $n \in \Bbb{N}$)?  \begin{align}
    D(x^2) &= D(x \cdot x) = D(x) \cdot x + x \cdot D(x) = 2 x D(x), \text{ and}  \\
    D(x^n) &= D(x \cdot x^{n-1}) = \dots = n x^{n-1} D(x) \text{.}
\end{align}  So, for any polynomial, $f$, $D(f) = f' D(x)$.  (Where we use the prime to denote "the usual derivative".  You introduced an alternative usage in your Question, but since you didn't subsequently use it, I'm stealing it to use here.)
What if we decide $D(x) = x^2$?  If $f$ and $g$ are polynomials, $$
    D(f \cdot g) = D(f) \cdot g + f \cdot D(g) = f' x^2 g + f g' x^2 = (fg)' D(x) \text{.}
$$  This example is warm-up for:  what if we decide $D(x) = \phi(x)$ for any function $\phi$?  If $f$ and $g$ are polynomials, $$
    D(f \cdot g) = D(f) \cdot g + f \cdot D(g) = f' \phi g + f g' \phi = (fg)' D(x) \text{.}
$$
So it appears that you can use any $\phi \in \mathcal{F}(\Bbb{R})$ as the image of $x$.  The image of $\mathcal{D}(\Bbb{R})$ will just be a "scaled and rotated" version of the normal derivative operator where the preimage of $0$ is the constants (and everything else, if you chose $\phi=0$) and the preimage of $\phi$ is $x$ and the rest of the image of $\mathcal{D}(\Bbb{R})$ falls out of this one choice.
