 Solution

Take the Fourier transform convention $\tilde{y}(\omega )={\int }_{\mathbb{R}}y(t)e^{-i\omega t}dt$, with inverse factor $e^{i\omega t}/(2\pi )$. Then $\widetilde{y^{(n)}}=(i\omega {)}^n\tilde{y}$, classically when the boundary terms vanish or in the sense of tempered distributions.

Multiply the given frequency relation by its denominator and then by $-i$:Recognizing the four derivative multipliers givesThe convention is stated explicitly because reversing the Fourier exponential changes intermediate frequency signs. The distributional form will matter at the jump of $f$ in part (b).