I have a function $x(t)$ that I'm multiplying with $\frac{d}{dt}\delta(t-kT)$ I know the property that $\frac{d}{dt}\delta(t-kT) = -\frac{\delta(t-kT)}{t-kT}$, and if I use that: $x(t)\frac{\delta(t-kT)}{kT-t} = x(kT)\frac{\delta(t-kT)}{kT-t}$
I'm not sure what this function looks like when $t=kT$, since the denominator is infinite, and $\delta(t-kT)$ is also of infinite height at $t=kT$ Is it just undefined, or does the delta function have some other property that I'm missing that makes it a defined function of some sort.
PS: I'm tagging this as signal processing since I encountered the function while doing signal processing stuff.
This is not an answer to your question, but I'm answering here so I can see what I'm TeX'ing.
I'm concerned that you are pursuing an incorrect approach. The identity $$\delta^\prime(t-kT) = -{\delta(t-kT)\over t-kT}$$ is false. To see this, integrate both sides of the identity against $e^{-t^2}$. The left-hand side yields $$\int_{-\infty}^\infty{\delta^\prime(t-kT)\, e^{-t^2}\,dt} =2kTe^{-(kT)^2}$$ while the right-hand side yields $$\int_{-\infty}^\infty{{\delta(t-kT)\over t-kT}\, e^{-t^2}\,dt} ={\rm ??}$$
Consulting the link you provided, I suspect you got $\delta^\prime(t-kT) = -{\delta(t-kT)\over t-kT}$ by dividing both sides by $x$ in the true identity $$x\delta^\prime(x) = -\delta(x)\,.$$ But that leads to trouble here.
Consider that if we could do such operations freely with delta functions, then from the true identity $x\delta(x) = 0$ we could derive the false identity $\delta(x) = 0$ by dividing both sides by $x$.