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 $F(s)$?

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    $\begingroup$ Fourier transform commutes with linear operators. Derivation is a linear operator. Game over. $\endgroup$
    – dohmatob
    Commented Nov 11, 2022 at 13:18

3 Answers 3


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)$

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    $\begingroup$ A very good answer indeed. +1 $\endgroup$ Commented Nov 9, 2014 at 18:59
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    $\begingroup$ @leonbloy Why exactly can we move the derivative inside the integral (apply Leibniz rule)? $\endgroup$
    – Konstantin
    Commented Mar 5, 2017 at 12:17
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    $\begingroup$ @Atsina In general, if you have $g(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} G(\omega) \, e^{i \omega t} d\omega $, then you know that $G(\omega)$ is the Fourier transform of $g(t)$ $\endgroup$
    – leonbloy
    Commented Sep 10, 2018 at 2:16
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    $\begingroup$ @leonbloy How 'w' came out of the integral as we are integrating with respect to 'dw'? Can anyone elaborate, please? $\endgroup$
    – UJM
    Commented Feb 13, 2021 at 10:24
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    $\begingroup$ I hate to be that guy, but doesn't your answer rely on the fact that the Fourier inversion formula holds for $f$? Which means you are assuming extra conditions on $f$? Like you are saying that $F$ is integrable (which it doesn't have to be in general). $\endgroup$
    – Not Euler
    Commented Sep 16, 2021 at 16:58

The Fourier transform of the derivative is (see, for instance, Wikipedia) $$ \mathcal{F}(f')(\xi)=2\pi i\xi\cdot\mathcal{F}(f)(\xi). $$


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 $\mathbb{R}$.)

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    $\begingroup$ Why the first term must vanish? I think we need the additional condition $\lim_{t\to\infty}f(t)=0$ to guarantee this. $\endgroup$
    – Xiang Yu
    Commented Feb 27, 2016 at 3:51
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    $\begingroup$ Since $f \in L^1 \cap C^1$ f is continous and integrable, and must tend to zero when t tends to infinity...? $\endgroup$
    – user202542
    Commented Jun 29, 2016 at 17:48
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    $\begingroup$ The limit need not exist, although if it exists it must be zero. There are smooth, i.e., $C^\infty$, $L^1$ functions that do not tend to zero as $x \to \infty$. For an example, just make smooth "spikes" of height 1 at each integer $n$, such that the spike at $n$ has width $2^{-n}$. The limit must be zero, however, if you replace "continuous" with "uniformly continuous." $\endgroup$
    – Zach
    Commented Jan 12, 2017 at 19:22
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    $\begingroup$ Where did we ever assume that $f$ is absolutely integrable, or was that an assumption appended as a bandage? $\endgroup$ Commented Dec 6, 2017 at 2:50
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    $\begingroup$ It seems that absolute integrability can NOT imply that $f$ vanishes at the infinity. See Did's answer here: math.stackexchange.com/questions/108191/… $\endgroup$
    – Sam Wong
    Commented Sep 18, 2018 at 2:53

One could derive the formula via dual numbers and using the time shift and linearity property of the Fourier transform. If $f$ is analytic around $x \in \mathbb{C}$, then (using $\varepsilon$ as the imaginary unit of dual numbers and assuming $y \in \mathbb{C} \setminus \left\{ 0 \right\}$): \begin{align*} f'\left( x \right) &= \frac{f\left( x \right) - f\left( x + y \cdot \varepsilon \right)}{y \cdot \varepsilon}\\ \mathcal{F}_{x}\left[ f'\left( x \right) \right]\left( x \right) &= \frac{F\left( x \right) - e^{2 \cdot \pi \cdot i \cdot y \cdot \varepsilon \cdot x} \cdot F\left( x \right)}{y \cdot \varepsilon} = \frac{F\left( x \right) - \left( 1 + 2 \cdot \pi \cdot i \cdot y \cdot \varepsilon \cdot x \right) \cdot F\left( x \right)}{y \cdot \varepsilon}\\ \end{align*} $$\fbox{$\mathcal{F}_{x}\left[ f'\left( x \right) \right]\left( x \right) = 2 \cdot \pi \cdot i \cdot x \cdot F\left( x \right)$}$$

Or more general: $$\fbox{$\mathcal{F}_{x}\left[ f^{\left( n \right)}\left( x \right) \right]\left( x \right) = \left( 2 \cdot \pi \cdot i \cdot x \right)^{n} \cdot F\left( x \right)$}$$


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