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I'm trying to take the derivative of: $\frac{-1}{6}(e^{-3t}-1) u(t)$

The $u(t)$ is the step response. So the answer I get is by just doing product rule: $\frac{1}{2}e^{-3t}u(t)-\frac{1}{6}e^{-3t}\delta(t)$ but wolfram gets a different answer.

Why does: $-\frac{1}{6}e^{-3t}\delta(t)$ goes to $0$?

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    The product rule still holds for distributions (which is what the delta function really is), so yes.2012-10-25

1 Answers 1

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Because "it has a bug"?

In my MMA (v8.0,Linux-x64),

 D[-(1/6) (E^(-3 t) - 1) HeavisideTheta[t], t] 

outputs

 -(1/6) (-1 + E^(-3 t)) DiracDelta[t]                            + 1/2 E^(-3 t) HeavisideTheta[t]