Let $s \ge 1$ and $n \ge 1$. Define $\langle\nabla\rangle^s$ as the Fourier multiplier with symbol $\langle\xi\rangle^s := (1 + |\xi|^2)^{s/2}$, so that $\widehat{\langle\nabla\rangle^s f}(\xi) = \langle\xi\rangle^s \hat{f}(\xi)$. For $f \in H^s(\mathbb{R}^n) \cap W^{1,\infty}(\mathbb{R}^n)$ and $g \in H^{s-1}(\mathbb{R}^n) \cap L^\infty(\mathbb{R}^n)$, the commutator $[\langle\nabla\rangle^s, f]g := \langle\nabla\rangle^s(fg) - f\,\langle\nabla\rangle^s g$ satisfies \begin{align*} \|[\langle\nabla\rangle^s, f]g\|_{L^2} &\le C_s\bigl(\|\nabla f\|_{L^\infty}\,\|g\|_{H^{s-1}} + \|f\|_{H^s}\,\|g\|_{L^\infty}\bigr), \end{align*} where $C_s > 0$ depends only on $s$ and $n$.