I know that $\widehat{1}=\delta$ with $\delta$ dirac delta function, and 1 tempered distribution.
Now, $\delta=\int_{\mathbb{R}^n}e^{-i\xi x}1dx\Rightarrow \delta'=\int_{\mathbb{R}^n}-ixe^{-i\xi x}dx=-i\widehat{f}$ with $f(x)=x$ The above is not rigorous at all. How can I rigorously proves this?
Actually it's easy to show that the formula $\widehat{xu} = i\hat{u}'$ is valid for a distribution $u$.
The Fourier transform $\hat{u}$ of a distribution $u$ is defined by $\langle \hat u, \varphi \rangle = \langle u, \hat \varphi \rangle$ for every test function $\varphi.$ This implies $$ \langle \widehat{xu}, \varphi \rangle = \langle xu, \hat\varphi \rangle = \langle u, x\hat\varphi \rangle = \langle u, -i\widehat{\varphi'} \rangle = -i \langle \hat u, \varphi' \rangle = i \langle \hat u', \varphi \rangle . $$ Thus, $\widehat{xu} = i\hat{u}'.$ Taking $u=1$ gives $\widehat{x} = i\delta'.$