Heaviside Function Derivative

Question

I came across this link that proves how the derivative of the heaviside function is the delta function, but I would like to ask whether -H'(-x) = $\delta$(x) in a distributional sense of course. Also, is [x$\delta$(x)]' = $\delta$(x) wrong (or right?) according to Wolfram on equation [17]? I am perhaps wrong as I am unfamiliar or incorrectly perceived the notations on Wolfram. Thankyou.

Answer 1

Since $H' = \delta$ we have $-H'(-x) = -\delta(-x) = -\delta(x).$ But perhaps you mean $-(H(-x))'$ (with a bit of abuse of notation)? For that we get $-(H(-x))' = -(-\delta(-x)) = \delta(-x) = \delta(x).$

Since $x\delta = 0$ we of course have $(x\delta)' = 0.$ However, $x\delta' = (x\delta)' - (x')\delta = 0' - 1\delta = -\delta.$