# How to differentiate the product signal $f(t)\theta(t)$, where $\theta(t)$ is Heaviside's unit step function?

What is the derivative (in the engineer's sense) of the causal function $f(t)\theta(t)$, where $\theta$ is the Heaviside unit step function?

I've seen the formula $f'(t)\theta(t)+f(0)\delta(t)$, where $\delta$ is Dirac's delta. This looks like a kind of "product rule": differentiating the product gives $f'\theta+f\theta'$, but $\theta'$ is $\delta$, and $\color{blue}{f(t)\delta(t)=f(0)\delta(t)}$.

If this is right, I don't understand the following argument, from the solutions manual to Oppenheim and Wilsky's Signals and Systems. The solutions manual says the derivative of the function $2e^{-3t}\theta(t-1)$ is

$$-6e^{-3t}\theta(t-1)+\color{red}{2}\delta(t-1)$$

It's the second term I don't understand. Using the "product rule" heuristic, the second term should be $2e^{-3t}\delta(t-1)$, which using the blue formula above gives $\color{red}{2e^{-3}}$ times the delayed delta function, not just twice the delayed delta function.

Is the solutions manual wrong?

(Cross-posted on MSE.)

• If you haven't made any typos, then the solutions manual is wrong and you are right. I've also seen a number of wrong results (mostly typos) in the solutions manual for the mentioned book. So this could be just another one. – Fat32 Feb 28 '17 at 23:09
• Please avoid cross-posting on multiple SE sites, it is not permitted. – Gilles Mar 1 '17 at 7:52

What I want to add here is that the product rule for distributions is by no means some "engineering heuristic" or some dubious magic, but it can be proved in a rigorous way (see e.g. this document, p. 5) by treating the combined function $f(t)\theta(t)$ as a distribution. I point this out because you don't seem to be sure about this (you mention: kind of "product rule", "product rule" heuristic).