# Fourier Transform of Derivative

Consider a function $f(t)$ with Fourier Transform $F(s)$. So

What is the Fourier Transform of $f'(t)$? Call it $G(s)$.So

Would we consider $\frac{d}{ds} F(s)$ and try and write $G(s)$ in terms of $F(s)$?

$$f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) \, e^{i \omega t} d\omegaf(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) \, e^{i \omega t} d\omega$$
$$f'(t) = \frac{d}{dt}\!\left( \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) \, e^{i \omega t} d\omega \right)= \frac{1}{2\pi} \int_{-\infty}^{\infty} i \omega \, F(\omega) \, e^{i \omega t} d\omegaf'(t) = \frac{d}{dt}\!\left( \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) \, e^{i \omega t} d\omega \right)= \frac{1}{2\pi} \int_{-\infty}^{\infty} i \omega \, F(\omega) \, e^{i \omega t} d\omega$$
Hence the Fourier transform of $$f'(t)f'(t)$$ is $$i \omega \, F(\omega) i \omega \, F(\omega)$$