I recently came across this paper that at first sight seems it could be relevant to your question: degruyter.com/view/j/math.2019.17.issue-1/math-2019-0006/…

A stronger notion of positivity for a polynomial $p(x)$ is that $|p(z)|\leq p(|z|)$ for all complex $z$. Assuming $p(0)\neq 0$ and the $\gcd$ of the exponents appearing in $p(x)$ to be $1$, an even stronger version is that $|p(z)|< p(|z|)$ for all complex $z$ not on the positive real axis. An example of such a polynomial with a negative coefficients is $p(x) = 2+2x-x^2+2x^3+2x^4$, while the polynomial $1-x+x^2$ does not satisfy the weaker version.

The way I look at it, $d/dx$ is the operator that takes the derivative of the one-variable function that follows it, and the symbol $x$ is (or should be) irrelevant. So it just operates on the space of all differentiable one variable functions. I think that when we write $\frac{d}{dx} x^2$ the confusing part is the $x^2$, that should be written as $x\mapsto x^2$. Then $\frac{d}{dx} (x\mapsto x^2)$ and $\frac{d}{dt}(x\mapsto x^2)$ give the same answer, $(x\mapsto 2x)$, while $\frac{d}{dx}(t\mapsto x^2)$ and $\frac{d}{dt}(t\mapsto x^2)$ are both $0$. But I know this clashes with common usage...