Derivative of unit step function

[shinobi20]

Homework Statement

Show that δ(x-x') = d/dx Θ(x-x')

Homework Equations

∫ f(x') δ(x-x') dx' = f(x)
Θ(x-x') vanishes if x-x' is negative and 1 if x-x' is positive

The Attempt at a Solution

I saw a relation of the δ function but I don't know why is it like that.

Integral of δ(x-x') from -∞ to x is 1 if x>x' and 0 if x<x'

I'm not sure how to start. Any suggestions?

 

[andrewkirk]

The statement is really only true for $x\neq x'$. Try to prove it first for $x<x'$ and then for $x>x'$.

Neither side has a value at $x=x'$. But what you can prove that can substitute for that (if one sets up the required structure based on treating them as distributions rather than functions) is that, for $a<x'<b:\ \int_a^b\delta(x-x')dx=\Theta(b)-\Theta(a)$.

If you can prove those three things, I think that's as close as one can get without getting deep into distribution theory.