Homomorphism under addition $\varphi :\mathbb{R}[x] \longrightarrow \mathbb{R}[x]$ where $\varphi(p(x))=\frac{d}{dx}(xp(x))$?

Question

I am wondering if $\varphi :\mathbb{R}[x] \longrightarrow \mathbb{R}[x]$ where $\varphi(p(x))=\frac{d}{dx}(xp(x))$ is really is a homomorphism. I can't think of a situation where it is not because I'm thinking that even $\mathbb{R}\subset \mathbb{R}[x]$. So if I had $p(x)=1, q(x)=2$ $$\varphi(p(x)+q(x))=\frac{d}{dx}((x)(1))+\frac{d}{dx}((x)(2))=1+2=\\=\varphi(p(x))+\varphi(q(x))$$

Is any of my reasoning wrong?

Thanks in advance for your help.

Answer 1

This is not a homomorphism of rings even if it is additive. In fact, for $p(x)=x, q(x)=x$, we have $$\varphi(p(x)q(x))=3x^2\neq \varphi(p(x)) \varphi(q(x))=2x\times 2x=4x^2.$$

Answer 2

The maps $p(x) \mapsto xp(x)$ and $p(x) \mapsto p'(x)$ are both additive and so is its composition $\varphi$.