Derivative of delta function not centered at x = 0

Question

I know that:

$x\delta'(x) = -\delta(x)$

Am I correct in thinking that in order to take $\frac{d}{dx}\delta(x-x_{0})$ I first need to set $u = x-x_{0}$, $du = dx$ so that:

$\frac{d}{dx}\delta(x-x_{0}) = \frac{du}{dx}\frac{d}{du}\delta(u) = -\frac{\delta(u)}{u}=-\frac{\delta(x-x_{0})}{x-x_{0}}$

It seems like a straight application of the chain rule, but I'm afraid I may be missing some subtlety of the delta function's properties.

Edit--to be more specific, I have a distribution function defined as such:

$g(x) = \int_{0}^{\infty}f(x')\delta(x-x')dx'$

and I'm trying to find $g'(x)$. Then:

$\frac{dg(x)}{dx} = \int_{0}^{\infty}f(x')\frac{d}{dx}\delta(x-x')dx' = -\int_{0}^{\infty}f(x')\frac{\delta(x-x')}{x-x'}dx' = \int_{0}^{\infty}f(x')\frac{\delta(x'-x)}{x'-x}dx'$

if I'm not mistaken.

Answer 1

After some related reading, I found what I was looking for here on the physics stack. I'll leave this post as a reference if anyone else has a similar question.