[proofplan]
We test both distributions against an arbitrary compactly supported smooth function $\phi \in \mathcal{D}(\Omega)$. The definition of [distributional derivative](/page/Distributional%20Derivative) turns $D_iT_f(\phi)$ into $-\int_\Omega f\,\partial_{x_i}\phi\,d\mathcal{L}^n$. Since $\phi$ has compact support inside $\Omega$, we extend the product $f\phi$ by zero to a compactly supported $C^1$ function on $\mathbb{R}^n$ and apply one-dimensional [integration by parts](/theorems/210) in the $x_i$ direction. The boundary term vanishes because of compact support, leaving exactly the [regular distribution](/page/Regular%20Distribution) associated to $\partial_{x_i}f$.
[/proofplan]
[step:Test the distributional derivative against a compactly supported smooth function]
Let $\phi \in \mathcal{D}(\Omega) = C_c^\infty(\Omega)$ be arbitrary. By the definition of the distributional derivative in the coordinate direction $i$,
\begin{align*}
D_iT_f(\phi) = -T_f(\partial_{x_i}\phi).
\end{align*}
Since $\partial_{x_i}\phi \in C_c^\infty(\Omega)$ and $f \in C^1(\Omega)$, the product $f\,\partial_{x_i}\phi$ is continuous with compact support in $\Omega$, so it is Lebesgue integrable on $\Omega$. Therefore
\begin{align*}
D_iT_f(\phi) = -\int_\Omega f(x)\partial_{x_i}\phi(x)\,d\mathcal{L}^n(x).
\end{align*}
[/step]
[step:Extend the compactly supported product to a global $C^1$ function]
Let $K := \operatorname{supp}\phi \subset \Omega$. Since $\phi$ has compact support in $\Omega$, $K$ is compact and there is an [open set](/page/Open%20Set) $V \subset \Omega$ such that $K \subset V$ and $\phi$ vanishes on $\Omega \setminus K$.
Define the function $u: \mathbb{R}^n \to \mathbb{R}$ by
\begin{align*}
u(x) = f(x)\phi(x)
\end{align*}
for $x \in \Omega$, and by $u(x)=0$ for $x \in \mathbb{R}^n \setminus \Omega$. Because $\phi$ vanishes on a neighbourhood of $\mathbb{R}^n \setminus \Omega$, this extension is a compactly supported $C^1$ function on $\mathbb{R}^n$. On $\Omega$, the product rule gives
\begin{align*}
\partial_{x_i}u(x) = \partial_{x_i}f(x)\phi(x) + f(x)\partial_{x_i}\phi(x).
\end{align*}
[guided]
The purpose of this step is to remove the boundary of $\Omega$ from the argument. We cannot directly integrate by parts on an arbitrary open set $\Omega$ without discussing its boundary. The compact support of $\phi$ lets us avoid that issue.
Let
\begin{align*}
K := \operatorname{supp}\phi.
\end{align*}
Because $\phi \in C_c^\infty(\Omega)$, the set $K$ is compact and satisfies $K \subset \Omega$. Hence $\phi$ is zero outside $K$, and in particular $\phi$ vanishes near every point of $\mathbb{R}^n \setminus \Omega$.
Define $u: \mathbb{R}^n \to \mathbb{R}$ by setting $u(x)=f(x)\phi(x)$ for $x \in \Omega$ and $u(x)=0$ for $x \notin \Omega$. This is well-defined because the formula involving $f$ is used only where $f$ is defined. The only possible issue is differentiability across $\partial\Omega$, but there is no actual transition to check: since $\phi$ vanishes on a neighbourhood of $\mathbb{R}^n \setminus \Omega$, the function $f\phi$ is already zero near the boundary. Thus the zero extension is $C^1$ on all of $\mathbb{R}^n$.
Inside $\Omega$, both $f$ and $\phi$ are $C^1$, so the classical product rule applies:
\begin{align*}
\partial_{x_i}u(x) = \partial_{x_i}f(x)\phi(x) + f(x)\partial_{x_i}\phi(x).
\end{align*}
Outside a neighbourhood of $K$, both $u$ and $\partial_{x_i}u$ vanish. Therefore $u$ is a compactly supported $C^1$ function on $\mathbb{R}^n$.
[/guided]
[/step]
[step:Integrate the global derivative and use vanishing boundary terms]
Choose $R>0$ such that $\operatorname{supp}u \subset (-R,R)^n$. Write $x = (x_i,x')$, where $x' \in \mathbb{R}^{n-1}$ denotes all coordinates except $x_i$. For each fixed $x' \in (-R,R)^{n-1}$, the one-variable function $t \mapsto u(t,x')$ is $C^1$ on $(-R,R)$ and vanishes at $t=-R$ and $t=R$. Hence
\begin{align*}
\int_{-R}^{R}\partial_{x_i}u(t,x')\,d\mathcal{L}^1(t) = u(R,x') - u(-R,x') = 0.
\end{align*}
Integrating this identity over $x' \in (-R,R)^{n-1}$ with respect to $\mathcal{L}^{n-1}$ gives
\begin{align*}
\int_{(-R,R)^n}\partial_{x_i}u(x)\,d\mathcal{L}^n(x) = 0.
\end{align*}
Since $\partial_{x_i}u$ is supported in $(-R,R)^n$, this is equivalently
\begin{align*}
\int_{\mathbb{R}^n}\partial_{x_i}u(x)\,d\mathcal{L}^n(x) = 0.
\end{align*}
[/step]
[step:Substitute the product rule and identify the regular distribution]
Using the product rule for $u=f\phi$ on $\Omega$ and the fact that both terms vanish outside $\Omega$ after extension by zero, we obtain
\begin{align*}
0 = \int_\Omega \partial_{x_i}f(x)\phi(x)\,d\mathcal{L}^n(x) + \int_\Omega f(x)\partial_{x_i}\phi(x)\,d\mathcal{L}^n(x).
\end{align*}
Rearranging gives
\begin{align*}
-\int_\Omega f(x)\partial_{x_i}\phi(x)\,d\mathcal{L}^n(x) = \int_\Omega \partial_{x_i}f(x)\phi(x)\,d\mathcal{L}^n(x).
\end{align*}
Combining this with the definition of $D_iT_f(\phi)$ yields
\begin{align*}
D_iT_f(\phi) = \int_\Omega \partial_{x_i}f(x)\phi(x)\,d\mathcal{L}^n(x).
\end{align*}
The right-hand side is $T_{\partial_{x_i}f}(\phi)$ by the definition of the regular distribution associated to $\partial_{x_i}f$. Since $\phi \in \mathcal{D}(\Omega)$ was arbitrary, the distributions agree:
\begin{align*}
D_iT_f = T_{\partial_{x_i}f}.
\end{align*}
[/step]
