Fourier Transform of Derivative Consider a function $f(t)$ with Fourier Transform $F(s)$. So $$F(s) = \int_{-\infty}^{\infty} e^{-2 \pi i s t} f(t) \ dt$$
What is the Fourier Transform of $f'(t)$? Call it $G(s)$.So $$G(s) = \int_{-\infty}^{\infty} e^{-2 \pi i s t} f'(t) \ dt$$
Would we consider $\frac{d}{ds} F(s)$ and try and write $G(s)$ in terms of $F(s)$?
 A: The Fourier transform of the derivative is (see, for instance, Wikipedia)
$$
\mathcal{F}(f')(\xi)=2\pi i\xi\cdot\mathcal{F}(f)(\xi).
$$
Why?
Use integration by parts:
$$
\begin{align*}
u&=e^{-2\pi i\xi t} & dv&=f'(t)\,dt\\
du&=-2\pi i\xi e^{-2\pi i\xi t}\,dt & v&=f(t)
\end{align*}
$$
This yields
$$
\begin{align*}
\mathcal{F}(f')(\xi)&=\int_{-\infty}^{\infty}e^{-2\pi i\xi t}f'(t)\,dt\\
&=e^{-2\pi i\xi t}f(t)\bigr\vert_{t=-\infty}^{\infty}-\int_{-\infty}^{\infty}-2\pi i\xi e^{-2\pi i \xi t}f(t)\,dt\\
&=2\pi i\xi\cdot\mathcal{F}(f)(\xi)
\end{align*}
$$
(The first term must vanish, as we assume $f$ is absolutely integrable on $\mathbb{R}$.)
A: A simpler way, using the anti-transform:
$$f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) \, e^{i \omega t} d\omega$$
$$f'(t) = \frac{d}{dt}\!\left( \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) \, e^{i \omega t} d\omega \right)= \frac{1}{2\pi} \int_{-\infty}^{\infty}   i \omega \, F(\omega) \, e^{i \omega t} d\omega$$
Hence the Fourier transform of $f'(t)$ is $ i \omega \, F(\omega)$
