Does any derivation of commutative algebra preserve its nil-radical?

Suppose $x\in N$, so that $x^n=0$ for some $n$. Then using the product rule for derivations many times, we see that $$ 0=D^n(x^n)=n! D(x)^n+Y, $$ where $Y$ is divisible by $x$. Therefore, $D(x)^{n^2}=(D(x)^n)^n$ is divisible by $x^n$, and therefore vanishes. Thus, $D(x)$ is nilpotent, and therefore $D(N)\subset N$.

Here is another cute argument (I don't remember where I learned it, I think it is folklore). Let $P\subset A$ be an arbitrary prime ideal. We claim it contains a $D$-stable prime ideal. For this, consider the mod $P$ Taylor map $$ f\colon A\to (A/P)[[t]] , a \mapsto \sum_{n\geq 0} \frac{D^n(a) \textrm{ mod } P}{n!} t^n.$$ A quick computation shows that $f$ is a ring map, and that for all $a\in A$ we have $f(D(a)) = \frac{d}{dt}(f(a))$. Therefore, the kernel $Q = \mathrm{ker}(f)$ is a $D$-stable ideal of $A$. Moreover, $Q$ is prime because $(A/P)[[t]]$ is a domain, and we have $Q \subset P$ because the constant term of $f(a)$ is $a \textrm{ mod } P$.

So every prime ideal of $A$ contains a $D$-stable prime ideal. Hence, the intersection of all prime ideals of $A$ equals the intersection of all $D$-stable prime ideals of $A$. But the former is the nilradical, and the latter is clearly $D$-stable.