Aobut delta and Heaviside function

[eljose]

we know that \delta(xa)=(1/a)\delta(x) if the dirac,s delta satisfies this then given the function H(ax) with H the Heaviside step function what relationship is there between H(ax) and H(x) with

\frac{dH}{dx}=\delta(x)

 

[qbert]

H(ax) = H(x).

 

[Crosson]

(H(ax))' = a H'(ax) = a \delta (ax) =\delta (x) = H'(x)