For the sake of completeness, here is the proof I suggested in the comments,
in some more detail.

**Lemma 1.** Let $\mathbf{k}$ be a commutative ring. Let $A$ be a $\mathbf{k}
$-algebra. Let $I$ be a two-sided ideal of $A$. Let $f:A\rightarrow A$ be a
derivation. Then, $f\left(  I^{n+1}\right)  \subseteq I^{n}$ for every
$n\in\mathbb{N}$.

Lemma 1 is Proposition 1.21 in my [*Collected trivialities on algebra
derivations*][1], where I prove it by straightforward induction on $n$.

Now, your claim follows from the following fact:

**Corollary 2.** Let $\mathbf{k}$ be a commutative ring. Let $A$ be a
$\mathbf{k}$-algebra. Let $I$ be a two-sided ideal of $A$. Let $f:A\rightarrow
A$ be a derivation. Let $a\in A$ and $n\in\mathbb{N}$ be such that $I^{n}a=0$.
Then, $I^{n+1}f\left(  a\right)  =0$.

*Proof of Corollary 2.* We must prove that $I^{n+1}f\left(  a\right)  =0$. In
other words, we must prove that $gf\left(  a\right)  =0$ for every $g\in
I^{n+1}$. So let us fix $g\in I^{n+1}$. We have $f\left(  \underbrace{g}_{\in
I^{n+1}}\right)  \in f\left(  I^{n+1}\right)  \subseteq I^{n}$ (by Lemma 1),
and thus $f\left(  g\right)  a\in I^{n}a=0$. In other words, $f\left(
g\right)  a=0$.

But $\underbrace{g}_{\in I^{n+1}=II^{n}}a\in I\underbrace{I^{n}a}_{=0}=0$, so
that $ga=0$.

Since $f$ is a derivation, we have $f\left(  ga\right)  =gf\left(  a\right)
+\underbrace{f\left(  g\right)  a}_{=0}=gf\left(  a\right)  $. Hence,
$gf\left(  a\right)  =f\left(  \underbrace{ga}_{=0}\right)  =f\left(
0\right)  =0$. This is exactly what we wanted to prove. Thus, Corollary 2 holds.

[1]: http://web.mit.edu/\symbol{126}darij/www/algebra/derivat.pdf