Proof involving Dirac Delta function

[ozone]

Prove that
$x \frac{d}{dx} [\delta (x)] = -\delta (x)$
this is problem 1.45 out of griffiths book by the way.

Homework Equations

I attempted to use integration by parts as suggest by griffiths using $f = x , g' = \frac{d}{dx}$
This yields $x [\delta (x)] - \int \delta (x)dx$

next I tried taking the derivative of both sides so that I would get my original
$x \frac{d}{dx} [\delta (x)] = x \frac{d}{dx}[\delta (x)]$

and so I'm back where I started. I have also tried using a definite integral so that

$\int _{-\infty }^{\infty }x \frac{d}{dx}[\delta (x)] \text{dx} = \int_{-\infty }^{\infty } x[\delta (x)] \, dx - \int_{-\infty }^{\infty } [\delta (x)] \, dx$

but we know that $\int_{-\infty }^{\infty } x[\delta (x)] \, dx = 0$ so our equation simplifies.
However this didn't get me any closer to solving the problem either.

I also have the second part of the problem regarding the step function which is defined as
$\theta (x) \begin{array}{ll} \{ & \begin{array}{ll} 1 & x>0 \\ 0 & x\leq 0 \\ \end{array} \\ \end{array}$ Show that $\frac{d}{dx}\theta (x) =\delta (x)$

This something I can grasp by stating that

$\frac{\Delta \theta }{\text{\Delta x}}(x=0 )\longrightarrow 1/0, \text{ as } \text{\Delta x}\longrightarrow 0.$
which is clearly an undefined(or infinite) slope, and at every other point on this graph the derivative is going to be 0. However I want to know if this is sufficiently rigorous, and if it isn't what a step in the right direction might be.

 

[Dick]

For the first part you should integrate against a test function f(x). Show that integrating f(x)*x*δ'(x) is the same as integrating f(x)*(-δ(x)).

 

[ozone]

Alright thanks I will try that. Can you give some sort of reasoning for why this method works to get the answer? It just seems strange to me to introduce a dummy function.