# 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

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

H(ax) = H(x).

#### Crosson

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