How can you show that $f\delta′=f(0)\delta′−f′(0)\delta$ for a function f that is infinitely differentiable? Assume that $f$ is infinitely differentiable. Let $\delta$ be the (Dirac) delta functional.
I know that $f\delta = f(0)\delta$, but I'm not sure how to derive the equation  $f\delta′=f(0)\delta′−f′(0)\delta$.
 A: I'm assuming you mean the delta function(al) $\delta(x)$.
If so, you should use formal integration by parts.
$$\int g\times (f\delta') =\int -(fg)'\delta = \int -(f'(0)g(x)+g'(x)f(0))\delta = -f'(0) \int g\delta-f(0)\int g'\delta$$
Integrating the last term by parts and stripping off the $\int g\times$ gives the result.
A: You're missing an $f$ in the left hand side.
Compute $(f\delta)'$ in two different ways.
On one hand $(f\delta)' = f'\delta + f\delta' = f'(0)\delta + f\delta'$.
On the other hand $(f\delta)' = (f(0)\delta)' = f(0)\delta'$.
Hence $f\delta' = f(0)\delta' - f'(0)\delta$.
A: (I suppose that with $\delta$ you mean the Dirac function.)
Just evaluate $(f(t)\delta(t))'$ in two ways.
First, use Leibniz' rule:
$$
(f(t)\delta(t))'=f(t)\delta'(t)+f'(t)\delta(t)=f(t)\delta'(t)+f'(0)\delta(t).
$$
We also have
$$
(f(t)\delta(t))'=(f(0)\delta(t))'=f(0)\delta'(t).
$$
Since the two expressions must be equal, we have
$$
f(t)\delta'(t)=f(0)\delta'(t)-f'(0)\delta(t).
$$
A: you can also expand $f$ into Taylor series 
$f(t)=f(0)+f'(0)t + \frac{f''(0)}{2!}t^2+\cdots$ 
and consider 
$(t \delta(t))' = 0 = \delta(t) + t\delta'(t)$
So we have 
$t\delta'(t) =- \delta(t)$
all high order terms involving $t^n \delta'(t)=0, n\ge2$. 
Finally
$f\delta′=f(0)\delta′−f′(0)\delta$.
