Why is \delta'(y) = -\delta'(-y)?

[signalcarries]

Homework Statement

I'm trying to prove that $$\delta'(y)=-\delta'(-y)$$.

Homework Equations

The Attempt at a Solution

I'm having trouble getting the LHS and the RHS to agree. I've used a test function $$f(y)$$ and I am integrating by parts.

For the LHS, I have
$$\int_{-\infty}^{\infty} f(y)\delta'(y)dy = \int_{-\infty}^{\infty} \frac{d}{dy}[f(y)\delta(y)]dy - \int_{-\infty}^{\infty} \delta(y)\frac{df(y)}{dy}dy = 0 - f'(0) = -f'(0)$$

For the RHS, I have
$$-\int_{-\infty}^{\infty} f(y)\delta'(-y)dy = \int_{\infty}^{-\infty} f(-t)\delta'(t)dt = -\int_{-\infty}^{\infty} f(-t)\delta'(t)dt = -\int_{-\infty}^{\infty} \frac{d}{dt} [f(-t)\delta(t)]dt + \int_{-\infty}^{\infty} \frac{df(-t)}{dt} \delta(t)dt = 0 + \int_{-\infty}^{\infty} \frac{df(-t)}{dt} \delta(t)dt = f'(0)$$.

I seem to be off by a minus sign, but I can't figure out where. Any help would be appreciated.

 

[Dick]

df(-t)/dt=-df(t)/dt at t=0.

 

[signalcarries]

Yes, I suppose it does. Thanks!

 

[the_amateur]

df(-t)/dt=-df(t)/dt at t=0.

A proof of the above statement would be more helpful.

 

[Dick]

A proof of the above statement would be more helpful.

Chain rule.