We consider the Heaviside function $H(x)$.
$H'(0)$ doesn't exist. The derivative exists if we define $H$ as a distribution
$$H: \phi \to \int_{-\infty}^{+\infty} H(x) \phi(x) dx= \int_0^{+\infty} \phi(x) dx$$
$\partial{H}$ is a function $\partial{H}: \phi \to \langle \partial{H}, \phi \rangle=- \langle H, \phi' \rangle=-\int_0^{+\infty} \phi'(x) dx=\phi(0)= \langle \delta, \phi \rangle$
Thus, $H'(x)=\delta(x)$.
If $f$ is smooth, I want to show that $(fH)'=f(0) \delta+ H f'$.
That's what I have tried so far:
$$(fH)(x)=\left\{\begin{matrix} f(x) &, x>0 \\ 0 &, x \leq 0 \end{matrix}\right.$$
We consider $fH$ as a distribution $$fH: \phi \to \int_{-\infty}^{+\infty} (fH)(x) \phi(x) dx=\int_0^{+\infty} f(x) \phi(x) dx$$
Then $$(fH)': \phi \to \langle (fH)', \phi\rangle=- \langle fH, \phi' \rangle=-\int_{0}^{+\infty} f(x) \phi'(x) dx=-f(0) \phi(0)+ \int_{0}^{+\infty} f'(x) \phi(x) dx $$
How can we continue in order to show that the latter is equal to $f(0) \delta+ H f'$ ?
Actually, you have $$ -\int_{0}^{+\infty} f(x) \phi'(x) dx={\color{red}+}f(0) \phi(0)+ \int_{0}^{+\infty} f'(x) \phi(x) dx=f(0) \phi(0)+ \langle Hf',\phi\rangle, $$ that is, $$ -\int_{0}^{+\infty} f(x) \phi'(x) dx=\langle f(0) \delta+ Hf',\phi\rangle. $$