Fourier Transform of derivative of $xf(x)$? What is the Fourier transform of $$\frac{\partial}{\partial x}(xf(x))$$
Shall I separate the product first? Then what is the Fourier transform of $$x\frac{\partial}{\partial x}f(x)$$
 A: Using this definition of the Fourier Transform
$$F(s) = \mathscr{F}\left\{f(x)\right\} = \int_{-\infty}^\infty f(x) e^{-2\pi i sx} \space dx$$
and these two theorems
$$\mathscr{F}\left\{\dfrac{d}{dx}f(x)\right\} = 2\pi i sF(s)$$
$$\mathscr{F}\left\{-2\pi i xf(x)\right\} = \dfrac{d}{ds}F(s)= F'(s)$$
one can derive the answer
$$\begin{align*}\mathscr{F}\left\{\dfrac{d}{dx}\left(xf(x)\right)\right\} &= 2\pi i s\mathscr{F}\left\{xf(x)\right\} \\
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&= - s\mathscr{F}\left\{-2\pi i xf(x)\right\} \\
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&= -s\dfrac{d}{ds}F(s)\\
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&= -s F'(s)
\end{align*}$$
Update to add a proof of the 2nd theorem
$$\begin{align*}\mathscr{F}^{-1}\left\{\dfrac{d}{ds}F(s)\right\} &= \int_{-\infty}^\infty {\left[\dfrac{d}{ds}F(s)\right]e^{2\pi i xs}}ds \\
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&= \int_{-\infty}^\infty {\lim_{\Delta s \to 0}{\dfrac{F(s+\Delta s) - F(s)}{\Delta s}}e^{2\pi i xs}}ds \\
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&= \lim_{\Delta s \to 0}{\dfrac{1}{\Delta s}\left[\int_{-\infty}^\infty {F(s+\Delta s)e^{2\pi i xs}}ds -\int_{-\infty}^\infty {F(s)e^{2\pi i xs}}ds\right]}\\
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&= \lim_{\Delta s \to 0}{\dfrac{1}{\Delta s}\left[\int_{-\infty}^\infty {F(s')e^{2\pi i x(s'-\Delta s)}}ds' -f(x)\right]}\\
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&= \lim_{\Delta s \to 0}{\dfrac{1}{\Delta s}\left[e^{-2\pi i x\Delta s}\int_{-\infty}^\infty {F(s')e^{2\pi i xs'}}ds'-f(x)\right]}\\
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&= \lim_{\Delta s \to 0}{\dfrac{e^{-2\pi i x\Delta s}f(x)-f(x)}{\Delta s}}\\
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&= \lim_{\Delta s \to 0}{\dfrac{e^{-2\pi i x(0+\Delta s)}-e^{-2\pi i x0}}{\Delta s}f(x)}\\
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&= \left(\dfrac{d}{ds}e^{-2\pi i xs}\right)\biggr|_{s=0} \space f(x)\\
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\mathscr{F}^{-1}\left\{\dfrac{d}{ds}F(s)\right\} &= -2\pi i x f(x)\\
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\dfrac{d}{ds}F(s) &= \mathscr{F}\left\{-2\pi i x f(x)\right\}\\
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\end{align*}$$
