(I'm following these notes by Vadim Kaplunovsky titled "Feynman Propagator of a Scalar Field," specifically asking about equation 10.)
When calculating the time derivative of the Feynman scalar propagator, we get a term $\delta(x^0-y^0)\times\langle 0|[\phi(x),\phi(y)]|0\rangle$. According to Kaplunovsky, the delta function forces the term to be zero at unequal times but the term is zero at equal times also because then the $\phi$'s commute. However, if one just naively plugs in $x^0=y^0$ into the equation, we get an indeterminate form $\infty\times0$. I realize this doesn't make sense because $\delta$ is a distribution, not a function, but is there a mathematical way to show that this term yields zero?