I need to prove the following identities regarding the Dirac delta function:
$$\text{1.) }\delta^{(n)}(-x) = (-1)^n\delta^{(n)}(x) \hspace{40pt}\text{2.) } x\delta^{(n)}(x) = -n\delta^{(n-1)}(x)$$
and one hint I'm given is to somehow use this property: $$\int_{-\infty}^{\infty}f(x)\delta^{(n)}(x)dx = (-1)^nf^n(0)$$
Any hints on how to go about doing this? I'm particularly at a loss regarding how to apply the given property since none of the given involves $f(x)\delta^{(n)}(x)$ save for the left-hand side of 2.) where $f(x) = x$
I don't know how to prove the first one, because in my opinion it follows right from the definition of changing variables in generalized functions. Let's prove second one. $$ (x \delta^{(n)}, \, \varphi) = (\delta^{(n)}, \, x \varphi) = (-1)^{n} (\delta, \, (x \varphi)^{(n)}) = (*) $$
By Leibnitz formula $$(x \varphi)^{(n)} = \sum \binom{n}{m} x^{(m)} \varphi^{(n-m)} = x \varphi^{(n)} + \binom{n}{1} x' \varphi^{(n-1)} = x\varphi^{(n)} + n \varphi^{(n-1)}.$$
Therefore,
$$ (*) = (-1)^{n} (\delta, \, x \varphi^{(n)}) + (-1)^{n} (\delta, \, n \varphi^{(n-1)}) = 0 + (-n \delta^{(n-1)}, \, \varphi). $$
So, $$ x\delta^{(n)} = -n \delta^{(n-1)}. $$
N.B. $(\cdot, \, \cdot)$ is a common notation for generalized functions. If you're not familiar with it, you can treat it as a scalar product in $L_2$:
$$ (f, \, g) = \int f(x) g(x) \, dx $$
Notice the property $(f', \, g) = - (f, \, g)$ (integration by parts!).