Fourier Transform of derivative of $xf(x)$?

Question

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 $$x\frac{\partial}{\partial x}f(x)$$

Answer 1

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

$$\begin{align*}\mathscr{F}\left\{\dfrac{d}{dx}\left(xf(x)\right)\right\} &= 2\pi i s\mathscr{F}\left\{xf(x)\right\} \\ \\ &= - s\mathscr{F}\left\{-2\pi i xf(x)\right\} \\ \\ &= -s\dfrac{d}{ds}F(s)\\ \\ &= -s F'(s) \end{align*}$$

Update to add a proof of the 2nd theorem $$\begin{align*}\mathscr{F}^{-1}\left\{\dfrac{d}{ds}F(s)\right\} &= \int_{-\infty}^\infty {\left[\dfrac{d}{ds}F(s)\right]e^{2\pi i xs}}ds \\ \\ &= \int_{-\infty}^\infty {\lim_{\Delta s \to 0}{\dfrac{F(s+\Delta s) - F(s)}{\Delta s}}e^{2\pi i xs}}ds \\ \\ &= \lim_{\Delta s \to 0}{\dfrac{1}{\Delta s}\left[\int_{-\infty}^\infty {F(s+\Delta s)e^{2\pi i xs}}ds -\int_{-\infty}^\infty {F(s)e^{2\pi i xs}}ds\right]}\\ \\ &= \lim_{\Delta s \to 0}{\dfrac{1}{\Delta s}\left[\int_{-\infty}^\infty {F(s')e^{2\pi i x(s'-\Delta s)}}ds' -f(x)\right]}\\ \\ &= \lim_{\Delta s \to 0}{\dfrac{1}{\Delta s}\left[e^{-2\pi i x\Delta s}\int_{-\infty}^\infty {F(s')e^{2\pi i xs'}}ds'-f(x)\right]}\\ \\ &= \lim_{\Delta s \to 0}{\dfrac{e^{-2\pi i x\Delta s}f(x)-f(x)}{\Delta s}}\\ \\ &= \lim_{\Delta s \to 0}{\dfrac{e^{-2\pi i x(0+\Delta s)}-e^{-2\pi i x0}}{\Delta s}f(x)}\\ \\ &= \left(\dfrac{d}{ds}e^{-2\pi i xs}\right)\biggr|_{s=0} \space f(x)\\ \\ \mathscr{F}^{-1}\left\{\dfrac{d}{ds}F(s)\right\} &= -2\pi i x f(x)\\ \\ \dfrac{d}{ds}F(s) &= \mathscr{F}\left\{-2\pi i x f(x)\right\}\\ \\ \end{align*}$$