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.)