Good evening! I can't understand how to prove that $$\alpha(x)\delta'(x)=-\alpha'(0)\delta(x)+\alpha(0)\delta'(x).$$ I tried to use $$(Df,\phi)=-(f, D\phi),$$ also I used that $$(D(hf),\phi)=(h'f+hf',\phi)$$ but I got another result. Thanks for help.
It would be very good, if you advised some books.
The derivative of the Dirac delta works exactly through the integration-by-parts scheme you've mentioned, $$(Df,\phi)=-(f, D\phi),$$ i.e. $\delta'(x)$ is the generalized function which returns for arbitrary $f$ $$ \int f(x) \delta'(x) \mathrm dx = -f'(0). $$ Multiplying $\delta'(x)$ by some regular function is essentially equivalent to doing the same with $f(x)$, so that \begin{align} (f, \alpha \, \delta') & = \int f(x) \alpha(x)\delta'(x) \mathrm dx \\ & = -\frac{\mathrm d}{\mathrm dx}\left[ f(x)\alpha(x)\right]_{x=0} \\ & = - \alpha'(0) f(0) - \alpha(0) f'(0) \\ & = - \alpha'(0) \,(f,\delta) + \alpha(0)\, (f,\delta') \\ & = (f,- \alpha'(0)\,\delta + \alpha(0)\,\delta') \end{align} as required.