How to show this identity: $\delta'(-x) = -\delta'(x)$ I was wondering how to prove the following property regarding the derivative of the Dirac delta function: $$\delta'(-x) = -\delta'(x)$$
I try to do it by multiplying both sides by some function $f(x)$ and then integrating by parts like so:
$$\int_{-\infty}^{\infty}f(x)\delta'(-x)dx = \int_{-\infty}^{\infty}-f(x)\delta'(x)dx$$
$$f(x)\delta(-x)\rvert_{-\infty}^{\infty} - \int_{-\infty}^{\infty}f'(x)\delta(-x)dx = -[f(x)\delta(x)\rvert_{-\infty}^{\infty} - \int_{-\infty}^{\infty}f'(x)\delta(x)dx]$$
$$0 - \int_{-\infty}^{\infty}f'(x)\delta(-x)dx = 0 + \int_{-\infty}^{\infty}f'(x)\delta(x)dx$$
Then using the property that $\delta(-x) = \delta(x)$:
$$-\int_{-\infty}^{\infty}f'(x)\delta(x)dx = \int_{-\infty}^{\infty}f'(x)\delta(x)dx$$
And here's where I keep getting stuck due to the differences in signs. Is there a mistake in my integration, or is there some other property of the Dirac delta function that I'm just not utilizing?

$\delta(x)=\delta(-x)$, so the result follows from the chain rule.
