The second derivative of the Dirac delta function I want to output the following expression: $tδ^{''}\left(t\right)=-2δ^{'}\left(t\right)$
My attempt, which did not lead to success: $\int_{ }^{ }δ^{'}\left(t-s\right)f^{''}\left(t\right)dt=f^{''}\left(s\right)$
$-\int_{ }^{ }δ^{''}\left(t-s\right)f^{'}\left(t\right)dt=f^{''}\left(s\right)$
$f^{'}\left(t\right)=g^{'}\left(t\right)\left(t-s\right)$
$-\int_{ }^{ }δ^{''}\left(t\right)\cdot t\cdot g^{'}\left(t+s\right)dt=g^{'}\left(s\right)$
$g^{'}\left(s\right)=\int_{ }^{ }g^{'}\left(t\right)δ\left(t-s\right)$
$δ^{''}\left(t\right)t=-δ\left(t\right)$
I probably made a mistake somewhere
 A: This should be a standard sort of computation, I think simpler than what you wrote. Short of doing the induction that proves, analogously, that $t\delta^{(n)}=-n\cdot \delta^{(n-1)}$, the explicit computations for $n=1,2$ are adequate to suggest the full case. The notation may be slightly different than what is familiar to you, but I claim it is very economical, and accurate. :)
Namely, $$(t\delta')(f)\;=\;\delta'(tf)\;=\;-\delta({d\over dt}(tf))\;=\;-\delta(f+tf')\;=\;-\delta(f)+(0\cdot f'(0)=-\delta(f)$$
That is, $t\delta'=-\delta$. An example of the induction step is
$$
(t\delta'')(f) \;=\; \delta''(tf) \;=\; (\delta')'(tf)\;=\; -\delta'({d\over dt}(tf)) \;=\; -\delta'(f+tf') \;=\; -\delta'(f)-\delta'(tf')
$$ $$ \;=\; -\delta'(f)-(t\delta')(f')
\;=\; -\delta'(f)+\delta(f')
\;=\; -\delta'(f)-\delta'(f) \;=\; -2\cdot \delta'(f)
$$
A: *

*Start with this identity of distributions
$$ t\delta(t)~=~0.$$

*Differentiate
$$ \delta(t)+t\delta^{\prime}(t)~=~0.$$

*Differentiate again
$$ 2\delta^{\prime}(t)+t\delta^{\prime\prime}(t)~=~0.$$
