Suppose \(f\in \mathcal{S}(R^n)\). Then
\[ \mathcal{F}(\partial^{\alpha}f)=i^{\lvert \alpha\rvert}\xi^{\alpha}\mathcal{F}f. \]In words, the Fourier transform turns a differentiation operator into multiplication one.
Indeed, this statement is true for Sobolev functions , i.e. \(f\in W^{m,2}(\mathbb{R}^n)\).