The Fourier transform of f′(t) is :
$$\hat{f'}(\omega) = \int_{-\infty}^{+\infty}f'(t)e^{-i\omega t}dt = f(t)e^{-i\omega t}\bigg\vert_{-\infty}^{+\infty} +i\omega\int_{-\infty}^{+\infty}f(t)e^{-i\omega t}dt = i\omega\hat{f}(\omega)$$
So $f(t)e^{-i\omega t}\bigg |_{-\infty}^{+\infty}$ must be zero but I don't understand how... ?
It's necessary that $f$ has compact support in $\mathbb{R}$, so if $f$ has compact support $K\subset \mathbb{R}$, then
$$f(t)e^{-iwt}\bigg\vert^{+\infty}_{-\infty}=0,$$
cause $f(x)=0$ for $x\in\mathbb{R}\backslash K.$