I'm trying to find the fourier transform of $t\cdot f''$. The rules I've found that relates to this seems to be that for a function $f(t)$ and it's Fourier transform $F(\omega)$ the following holds: $\frac{d^nf(t)}{dt^n} = (i\omega)^nF(\omega)$ and $(-it)^nf(t) = \frac{d^nF(\omega)}{d\omega^n}$.
Does this mean that the Fourier transform is $(i\cdot t)^2 \cdot \frac{dF(t)}{d\omega}$?