= Solution
Write $F(\omega)=\widetilde f(\omega)$ to distinguish the <Fourier transform> from the original function. <Integration by parts>, using the stated decay, gives
$$
\widehat{f'}(\omega)=\big[f(t)e^{-i\omega t}\big]_{-\infty}^{\infty}+i\omega\int_{-\infty}^{\infty}f(t)e^{-i\omega t}\,dt=i\omega F(\omega).
$$
Applying the same argument to $f'$ gives $\widehat{f''}=i\omega\widehat{f'}=-\omega^2F$. Differentiation with respect to the frequency gives
$$
F'(\omega)=\int_{-\infty}^{\infty}(-it)f(t)e^{-i\omega t}\,dt=-i\widehat{tf}(\omega).
$$
Consequently
$$
\boxed{\widehat{f'}=i\omega F,\qquad\widehat{f''}=-\omega^2F,\qquad\widehat{tf}=iF'.}
$$
The classical steps require existence of the relevant integrals and justification for differentiation under the integral, for example sufficient integrability of $f$, its derivatives and $tf$. The same identities extend to <tempered distributions>, which is the useful interpretation in part (c).
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