# Derivative of Dirac Delta function

• LedPhoton

#### LedPhoton

Hello I'm trying to figure out how to evaluate(in the distribution sense)
$\delta'(g(x))$. Where $\delta(x)$ is the dirac delta function. Please notice that what I want to evaluate is not $\frac{d}{dx}(\delta(g(x)))$ but the derivative of the delta function calculated in g(x).
If anyone could post a proof, an idea to find the proof or a link it would be greatly appreciated!

$\delta'$ is a linear operator on functions. Are you referring to its value when you pair it with g(x), as in
$$\langle \delta',\, g\rangle = \int_{\mathbb{R}} \delta'(x)g(x)\, dx \, ?$$
I'll assume you are. In that case, you use the integral notation above and then symbolically do integration by parts. Don't worry if it is not a well-defined operation because the answer you get is literally the definition of what you want.

http://en.wikipedia.org/wiki/Dirac_delta_function

No I'm sorry if I wasn't clear. I understand the value of
$\int \delta'(x)g(x)dx$
$\int \delta'(g(x))\phi(x)dx$
Where $\phi(x)$ is the test function. Here it is not immediately obvious to me how to integrate by parts. I thought about this(but I am unsure of whether it is correct):
Assume that g(x) is an invertible function with as many derivatives as necessary(to keep things simple for now), so we substitute $y = g(x)$ and get
$\int \delta'(y)\frac{\phi(g^{-1}(y))}{g'(g^{-1}(y))} dy$ Now I could integrate by parts and get
$-\int \delta(y)\frac{d}{dy}(\frac{\phi(g^{-1}(y))}{g'(g^{-1}(y))}) dy$
Do you think my reasoning is correct up to here?

The dirac delta is just a normal distribution who's standard deviation approaches 0. Take the derivative of the normal dist. then take the limit as stdev =>0. I'm not sure if that's a valid way to do the problem, but its what I would try.

Ok thanks, I'll try that

Your calculation looks right to me, and what Aero said makes sense too. As far as proof goes, I can't remember exactly how general the rules are for changing variables like that. A book like Friedlander would probably have it...

Check out the very last post here for a similar problem: