Well, yes, I should have said that the magnitude is 1 everywhere, but the argument is true: f is not defined at x=0. The derivative is zero on the entire domain of f. I guess that that is what Mathemaitca wants to say.

I am aware that the delta distribution can be considered to be the derivative of the Heaviside function.

http://mathworld.wolfram.com/DeltaFunction.htmlNote that the Heaviside function, defined as a piecewise constant function, is defined at x=0, so it is defined on the entire real line. (There is a way of defining as as a generalised function as well, which is a bit different.) BTW, I believe that Mathematica basically uses the piecewise constant function definition of the Heaviside function.

http://mathworld.wolfram.com/HeavisideStepFunction.htmlThe derivative of the Heaviside function is indeed the Dirac Delta:

D[HeavisideTheta[x], x]

(*DiracDelta[x]*)

I think that Mathematica is just very cautious about distinguishing between

x/(Sqrt[x^2])

and

D[2*HeavisideTheta[x] - 1, x]

(*2 DiracDelta[x]*)

is indeed twice the DiracDelta. But they are not equal because of the domains. See also my post below. In other words there is no "jump" anywhere in its domain. It is a bit like with a function which is defined on the rationals such that it is 0 for all rationals < Sqrt[2] and 1 for all rationals > Sqrt[2]. Over the rationals that function is continuous in spite of the obvious jump at Sqrt[2]. There is no point in the rationals where it actually is discontinuous. It's all a matter of the domains.

M.