Fourier transform of $g(x)=x\frac{\partial f}{\partial x}$? I have a problem with the Fourier transform of the function $g(x)=x\frac{\partial f}{\partial x}$. I need the transform to be itself a function of the Fourier transform of $f(x)$ and I don't know how to go about this problem.
I know that if $g(x)=\frac{\partial f}{\partial x}$, the FT would be $\hat{g}=is\hat{f}$. I also know that if $g(x)=xf(x)$, the FT would be $\hat{g}=i\frac{\partial {\hat{f}}}{\partial x}$. But what is the FT of $g(x)=x\frac{\partial f}{\partial x}$ as a function of $\hat{f}$? I would really appreciate if anyone could give any hints. Thank you. 

Let
$$\hat{g}(k) = \int_{-\infty}^{\infty} dx \: f'(x) e^{i k x} = -i k \hat{f}(k)$$
Then
$$\begin{align}\frac{d}{dk} \hat{g}(k) &= i \int_{-\infty}^{\infty} dx \: x f'(x) e^{i k x}\\ &=  -i \left [ k \frac{d}{dk} \hat{f}(k) + \hat{f}(k) \right ]\end{align}$$
Therefore
$$ \int_{-\infty}^{\infty} dx \: x f'(x) e^{i k x} = -\left [ k \frac{d}{dk} \hat{f}(k) + \hat{f}(k) \right ]$$
This of course assumes we can differentiate under the integral sign, which places restrictions on $f$ (i.e. that it vanish at least as $1/|x|$ as $x \rightarrow \pm \infty$).
