# Prove frequency derivative property of Fourier transform

If$$\mathcal{F}[x(t)]=X(ω)$$the duality property gives: $$\mathcal{F}\left[X(t)\right]=2\pi x(-ω)$$

I want to prove derivative in frequency$$\mathcal{F}[-jωx(t)]=\frac{dX(ω)}{dω}$$ using the duality property.

Maybe I'm confused as to how we use the dualty property. We have $$\mathcal{F}\left(\frac{dx}{dt}\right)=jωX(ω)$$ Now using duality $$\mathcal{F}\left[jtX(t)\right]=2\pi \frac{dx(-ω)}{dω}$$ How do I proceed now?

\begin{align}\frac{d \mathcal{F}(\omega)}{d\omega}&=\int_{-\infty}^{+\infty}f(t)\frac{d}{d\omega}\cdot e^{i\omega t}\,dt\\&=-i\int_{-\infty}^{+\infty}t.f(t)e^{i\omega t}\,dt\\&=-i \cdot\mathcal{IFT}\{t \,.f(t)\}\end{align}
$$\mathcal{F}\{t.f(t)\}=i.\frac{dF}{d\omega} \implies \mathcal{F}\{(-it).f(t)\}=\frac{dF}{d\omega}$$