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


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.)

  • $\begingroup$ 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. $\endgroup$ – Fat32 Feb 28 '17 at 23:09
  • $\begingroup$ Please avoid cross-posting on multiple SE sites, it is not permitted. $\endgroup$ – Gilles Mar 1 '17 at 7:52

The answer has already been given in Fat32's comment: if the solution manual says what you claim then it is wrong and you are right.

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).

Furthermore, also the fact that the (generalized) derivative of the step function equals the Dirac delta impulse can be shown in a rigorous way. So even if we use notation that seems appalling to most mathematicians, we engineers tread on pretty safe ground when it comes to the use of distributions as given in your example.

A very good introduction to distributions and singularity functions that is rigorous yet accessible for engineers can be found in The Fourier Integral and Its Applications by A. Papoulis.

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  • $\begingroup$ Thanks for the reference: yes, you're right to pick up on my guarded language, because I wasn't sure. My training is in math, not engineering, but I happen to be reading an engineering book, where nothing is done rigorously, and even useful facts like the product rule aren't named or discussed. $\endgroup$ – symplectomorphic Mar 1 '17 at 13:17

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