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

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Wolfram takes classical derivatives where they exist (and only those). The heaviside function is not differentiable at zero in the classical sense, so Wolfram doesn't bother taking the derivative there. What you are taking is a weak derivative, and I don't think Wolfram is capable of doing those. – Chris Janjigian Oct 25 '12 at 1:25
@Chris so then is my answer perfectly correct? – Charlie Yabben Oct 25 '12 at 1:27
1  
The product rule still holds for distributions (which is what the delta function really is), so yes. – Chris Janjigian Oct 25 '12 at 1:30

1 Answer

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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]
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