If $T$ is a distribution on $\mathcal{D}$, we define the derivative distribution $T'$ by
$$\forall \varphi \in \mathcal{D}, \quad T'(\varphi) = -T(\varphi')$$
Let $U$ be the function defined by
$$\begin{cases} U(x) = 1 & \text{if } x \geqslant 0 \\ U(x) = 0 & \text{if } x < 0 \end{cases}$$
Show that $T_U' = \delta_0$.