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) <br /> \begin{array}{ll}<br /> \{ &amp; <br /> \begin{array}{ll}<br /> 1 &amp; x&gt;0 \\<br /> 0 &amp; x\leq 0 \\<br /> \end{array}<br /> \\<br /> \end{array}<br /> Show that \frac{d}{dx}\theta (x) =\delta (x)

This something I can grasp by stating that

<br /> \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.