Prove the Dirac Delta Function satisfies $ x\frac{\mathrm{d} \delta(x)}{\mathrm{d} x} = -\delta(x) $ $ x\frac{\mathrm{d} \delta(x)}{\mathrm{d} x} = -\delta(x)$
I've been told that this answer involves integration by parts. I began like this:
$\int x\frac{\mathrm{d} \delta(x)}{\mathrm{d} x} = x\delta(x) - \int\delta(x)$
or
$ x\frac{\mathrm{d} \delta(x)}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(x\delta(x)) - \delta(x)$
So it seems that all I have to do is show that $x\delta(x) =0$ or $\frac{\mathrm{d} }{\mathrm{d} x}( x\delta(x)) =0$ 
EDIT: But  $x\delta(x) =0$ when $x=0$ and also when $x$ is any other number. so I just answered my own dumb question. 
Griffiths. Introduction to Electrodynamics. Third Edition page 49. Problem 1.45 part a. 
 A: Lets look at $(\delta'(x)x,\psi(x))$
Where $\psi(x)$ is a suitable test function and (,) denotes the inner product (remember that we can interpret an distribution as a linear functional over the space of test functions) then we can apply the general rule $(\delta', f)=-(\delta, f')$.
Furthermore we can push $x$, or more generally speaking any smooth function onto the testfunction (roughly speaking, because this gives another one) and so we get:
$$
(\delta'(x)x,\psi(x))=-(\delta(x),(x \psi(x))')=-(\delta(x),\psi(x)+x \psi'(x))=-(\delta(x),\psi(x))+\underbrace{(\delta(x),x\psi'(x))}_{=0}
$$
the last equation follows from the fact that $x\psi'(x)$ is again a testfunction with the propertiy $0\psi'(0)=0$ and therefore $(\delta,0)=0$
Maybe we should add the point, that every testfunction is bounded, so that my last line is really justified now :)
Edit:
To make this answer look at little bit more "physical" one replaces the inner product by $\int dx$ then everything should look a little bit more familiar... Furthermore just think about this test functions as functions which are as "well behaved" as necessary to allow all the manipilations
A: Everything you know about a distribution comes from how it acts on test functions. So if you want to prove anything about distributions, you should use an arbitrary test function, say $\phi$. In terms of physical intuition, I tend to think of a distribution as an object with some unknown properties and test functions as "observations" of the object, like the blind men and the elephant. YMMV of course.
The definition of the derivative of a distribution $f$ comes from integration by parts: $f'(\phi)(x) = - f(\phi')(x)$. Since integration by parts comes from the product rule and (you should prove this) distributional derivatives satisfy the product rule, we should suspect something like:
$$ \frac{d}{dx}(x\delta) = \delta + x\frac{d\delta}{dx}$$
So now, as you observe, $$(x\delta)(\phi)(x) = \int x\delta\phi\ dx = \int \delta x\phi\ dx = (x\phi)(0) = 0.$$
(I mostly answered this to add a bit of background and philosophy about distributions.)
