I know that the derivative of the Heaviside step function follows:
$${\partial \over {\partial {t_1}}}\theta \left( {{t_1} - {t_2}} \right) = \delta \left( {{t_1} - {t_2}} \right)$$
what about if the derivation with respect to t2? as follows:
$${\partial \over {\partial {t_2}}}\theta \left( {{t_1} - {t_2}} \right) = ??$$
Edit: I found the answer in
Kokhanovsky, Alexander A. Light scattering reviews 3: light scattering and reflection. Springer Science & Business Media, 2008
Equation (A.22): $${\partial \over {\partial {t_2}}}\theta \left( {{t_1} - {t_2}} \right) = - \delta \left( {{t_2} - {t_1}} \right)$$ and because of the Dirac delta function is even, we can set: $${\partial \over {\partial {t_2}}}\theta \left( {{t_1} - {t_2}} \right) = - \delta \left( {{t_1} - {t_2}} \right)$$
If we can justify that $$ {\mathrm{d}\over\mathrm{d} t}\theta(t) = \delta(t) $$ in some sense (not gonna open that can of worms), we can define $t=t_1 - t_2$ and use the chain rule to show that $$ {\partial\over\partial t_1}\theta(t) = {\mathrm{d}\theta\over\mathrm{d}t}{\partial t\over\partial t_1} = \delta(t) = \delta(t_1 - t_2) $$ and $$ {\partial\over\partial t_2}\theta(t) = {\mathrm{d}\theta\over\mathrm{d}t}{\partial t\over\partial t_2} = -\delta(t) = -\delta(t_1 - t_2). $$