$D: \Bbb R[x]\to \Bbb R[x]$ satisfies $D(fg)=D(f)g+fD(g), D(x)=1$.Can we show that $D$ is linear? 
$D: \Bbb R[x]\to \Bbb R[x]$ satisfies $D(fg)=D(f)g+fD(g), D(x)=1$.Can we show that $D$ is linear?

Clearly, we have $D(x^k)=kx^{k-1}$, $D(1)=0$. But these two properties can ensure that $D$ is linear?
 A: Define $\nu\colon\mathbb R[x] \to \mathbb N_0$ with formula $$\nu(f(x)) = \max\{ k\in\mathbb N_0 : x^k \text{ divides }f(x) \}$$
and $D\colon\mathbb R[x]\to \mathbb R[x]$ with formula $$D(f(x)) = \nu(f(x))x^{-1}f(x).$$
Note that $\nu(fg) = \nu(f) + \nu(g)$ and so $$D(fg) = \nu(fg)x^{-1}fg = \nu(f)x^{-1}fg + f\nu(g)x^{-1}g = D(f)g + f D(g)$$ and $D(x) = 1\cdot x^{-1} \cdot x = 1$. However, $D(x+1) \neq D(x) + D(1)$.
A: We have the following observation:

Proposition. Let $F$ be a field, and define the operator $D : F[x] \to F[x]$ as follows:

*

*For each irreducible monic polynomial $f$, assign an arbitrary value to $D(f)$.

*For any non-zero polynomial $f$ with the factorization $f = c \prod_k f_k$ into (not necessarily distinct) irreducible monic $f_k$'s and $c \in F^{\times}$, set $ D(f) = f \sum_k \frac{D(f_k)}{f_k} $.

Then $D$ is well-defined and solves the functional equation
$$ D(fg) = D(f)g + fD(g), \qquad f, g \in F[x]. \tag{FE1} $$

In light of this, we can come up with $D$ that is not the same as the differential operator $\frac{\mathrm{d}}{\mathrm{d}x}$ but still satisfies $D(x) = 1$.

Example. @Ennar's construction amounts to setting $D(x) = 1$ and $D(f) = 0$ for all the other irreducible monic $f$ in $\mathbb{R}[x]$.

So, how do we prove the claim? As it will turn out, the proposition becomes almost trivial if we consider the auxiliary operator $L$ defined by
$$ L(f) = \frac{D(f)}{f}. $$
Then the functional equation $\text{(FE1)}$ is equivalent to
$$ L(cf) = L(f) \qquad\text{and}\qquad L(fg) = L(f) + L(g) \tag{FE2} $$
for any $c \in F^{\times}$ and $f, g \in F[x]\setminus\{0\}$.
Now, the uniqueness factorization property of $F[x]$ allows us to regard $M = F(x)^{\times}/F^{\times}$ as a free abelian group with the generators being irreducible monics. So, any function $L$ defined on those generators of $M$ with values in $F(x)$ uniquely extends to a group homomorphism from the multiplicative group $M$ to the additive group $F(x)$, the fact being equivalent to $\text{(FE2)}$. Then establishing the rest is straightforward.
