We know that the product of the delta function and another function samples the latter function. That is, $$ \delta(t-\tau)f(t)=\delta(t-\tau)f(\tau) $$
Does the doublet function retain this same property? That is, does the following hold true: $$ \delta'(t-\tau)f(t)=\delta'(t-\tau)f(\tau) $$ My reasoning is that $\delta'(t-\tau)$ is only nonzero at $\tau$, and therefore the value of $f$ at any time other than $\tau$ does not matter, in the same manner as the delta function product.
One of this issues I have with this result though is that computing the Laplace transform of the doublet function does not seem to work out. $$ \mathcal{L}\{\delta'(t)\} = \int_{-\infty}^{\infty} \delta'(t)e^{-st}dt = \int_{-\infty}^{\infty} \delta'(t)e^{-s\times0}dt = \int_{-\infty}^{\infty} \delta'(t)dt = 0 $$ The final integral is zero since the doublet function is odd. However, we know that since the doublet is the derivative of the delta, its Laplace transform must be $\mathcal{L}\{\delta'(t)\}=s$.
I would appreciate if someone could offer some insight.