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, 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.

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, 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.