[proofplan] We verify the Leibniz rule by testing both sides against an arbitrary test function $\varphi \in \mathcal{D}(\Omega)$. The computation reduces to applying the definitions of the distributional derivative and multiplication by a smooth function, then invoking the classical Leibniz rule for the smooth functions $\psi$ and $\varphi$. [/proofplan] [step:Expand $\partial_j(\psi T)$ using the definitions of distributional derivative and smooth multiplication] Let $\varphi \in \mathcal{D}(\Omega)$. By the definition of the [distributional derivative](/page/Distributional%20Derivative), $(\partial_j(\psi T))(\varphi) = -(\psi T)(\partial_j \varphi)$. By the definition of multiplication of a distribution by a smooth function, $(\psi T)(\partial_j \varphi) = T(\psi \, \partial_j \varphi)$. Therefore \begin{align*} (\partial_j(\psi T))(\varphi) &= -T(\psi \, \partial_j \varphi). \end{align*} Apply the classical Leibniz rule to the smooth functions $\psi, \varphi \in C^\infty(\Omega)$: \begin{align*} \psi \, \partial_j \varphi &= \partial_j(\psi \varphi) - (\partial_j \psi)\varphi. \end{align*} Substituting into the expression for $T$: \begin{align*} -T(\psi \, \partial_j \varphi) &= -T(\partial_j(\psi \varphi)) + T((\partial_j \psi)\varphi). \end{align*} The first term equals $(\partial_j T)(\psi \varphi)$ by the definition of the distributional derivative of $T$, which in turn equals $(\psi(\partial_j T))(\varphi)$ by the definition of smooth multiplication. The second term equals $((\partial_j \psi) T)(\varphi)$ by the definition of smooth multiplication. Therefore \begin{align*} (\partial_j(\psi T))(\varphi) &= (\psi(\partial_j T))(\varphi) + ((\partial_j \psi) T)(\varphi). \end{align*} Since this holds for every $\varphi \in \mathcal{D}(\Omega)$, we conclude $\partial_j(\psi T) = (\partial_j \psi) T + \psi(\partial_j T)$ in $\mathcal{D}'(\Omega)$. [/step]