You mean the derivative of Sqrt[x^2]/x, not x/Sqrt[x^2], right?

As Marco said, mathematically this expression is undefined for x=0, and so is its derivative.

However, Mathematica simply takes the formal derivative (which results in 0) and does not look at for what values of x the expression is not defined.  This is a rather common issue in Mathematica, but I think that trying to be completely precise in these cases would turn out to be counterproductive.

Another example would be (2x-2)/(x-1).  This expression is equal to 2 for all x except for x=0 where it's undefined, hence its derivative should be undefined too.  But once again Simplify[(2x-2)/(x-1)] takes the practical route and gives 2, which is often the preferable result (although it's not strictly mathematically valid).  Simplify@D[(2x-2)/(x-1), x] is results 0 too.

It is good to be aware of this behaviour: often the generic result returned by Mathematica is not valid in a one or a few points (usually similar cases to the above).

Are you sure about the Dirac Delta?
The function you define (in the Reals?) is 1 everywhere apart from 0, where it is not defined. At that point (x=0) it is not defined, and hence has no derivative. I would think that Mathematica's answer that the derivative is 0 everywhere on its domain is quite correct.

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.html
Note 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.html

The derivative of the Heaviside function is indeed the Dirac Delta:

I think that Mathematica is just very cautious about distinguishing between 

and 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.