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I know that

$\frac{d}{dx} H(x) = \delta(x)$

but if the heaviside function is of two variables, what would the derivative be? I've searched but not found any discussions on this matter. I have an exercise where i need to calculate

$\frac{d}{dt} ( H(x-\xi+ct) - H(x-\xi-ct) )$, where $\xi, c$ are constant.

but as I said, I don't know how it works when there in two variables.

Thanks

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Actually, it does not matter that there are two variables. You are only asked to compute the derivative with respect to $t$, so you can (for this computation) assume $x$ to be a constant. Hence, you only have to apply the chain rule to get

\begin{align*} & \frac{d}{dt}( H(x-\xi + ct) - H(x - \xi - ct)) \\ &= H'(x - \xi + ct) \cdot (\frac{d}{dt} (x - \xi + ct)) - H'(x-\xi - ct) \cdot (\frac{d}{dt}(x - \xi - ct))\\ & = \delta(x - \xi + ct) \cdot c - \delta(x - \xi - ct) \cdot (-c)\\ & = c \cdot (\delta(x - \xi + ct) + \delta(x - \xi - ct)) \end{align*}

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  • $\begingroup$ Ooooh! I feel like such an idiot now, I was doing exactly what you wrote, But I forgot the minus sign of the inner derivative from the second term, therefore I was getting the wrong answer and hence thought that i was wrong about the derivation ;) Thanks alot though! @GenericNickname $\endgroup$ – desa Mar 1 '16 at 9:45

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