Differentiating Dirac delta with product rule

Question

I have here an equation. $$ h'(t_2) \delta(t_1 - t_2) = [h(t_2) - h(t_1)] \delta'(t_1 - t_2) $$

I checked the equality by integrating both sides with a test function. $$ \int d t_1 \phi(t_1) \ldots \to h'(t_2) \phi(t_2) \\ \int d t_2 \phi(t_2) \ldots \to h'(t_1) \phi(t_1) $$

Is this equation mathematically correct? I see a very similar derivation in this answer. If correct, can this kind of equation be derived using some sort of product rule? Checking using test functions may not be very practical for more complicated expressions. E.g. $$ h'(t_1) \delta(t_1 - t_3) = \int d t_2 h(t_2) [\delta'(t_1 - t_2) \delta(t_2 - t_3) - \delta(t_1 - t_2) \delta'(t_2 - t_3)] $$

Answer 1

For $\phi\in C_C^\infty$ and $h\in C^\infty$, we have

$$\begin{align} \langle h'_{t_2}\delta_{t_2},\phi \rangle&=\langle \delta_{t_2},h'_{t_2}\phi \rangle\\\\ &=h'(t_2)\phi(t_2)\tag1 \end{align}$$

and

$$\begin{align} \langle h_{t_2}\delta'_{t_2},\phi \rangle&=\langle \delta'_{t_2},h_{t_2}\phi \rangle\\\\ &=-h(t_2)\phi'(t_2)\tag2 \end{align}$$

and

$$\begin{align} \langle h\delta'_{t_2},\phi \rangle&=\langle \delta'_{t_2},h\phi \rangle\\\\ &=-h'(t_2)\phi(t_2)-h(t_2)\phi'(t_2)\tag3 \end{align}$$

Subtracting $(3)$ from $(2)$ yields $(1)$. And we are done!

Answer 2

Start with $h(t_1) \delta(t_1 - t_2) = h(t_2) \delta(t_1 - t_2)$. Differentiate both sides w.r.t. $t_1$ using the product rule $$ h'(t_1) \delta(t_1-t_2) + h(t_1) \delta'(t_1 -t_2) = h(t_2) \delta'(t_1-t_2) $$