The Fourier Derivative theorem states that
$ \widetilde{\dfrac{df(x)}{dx}} = ik\widetilde{f(k)}$
where the over-tilde denotes application of the Fourier transform.
My question is what if $f(x)$ is of the form $f(x) = g(x)\dfrac{dh(x)}{dx}$ and I want to apply the transform
$ \cal{F} \left[ {\dfrac{d}{dx} \left(g(x)\dfrac{dh(x)}{dx} \right)} \right] $
I was initially inclined to just apply the derivative theorem to each operator, however the fact that $g(x)$ is a function of x is causing me to doubt this is correct.
Do I have to apply the product rule first to give
$\dfrac{d}{dx} \left(g(x)\dfrac{dh(x)}{dx} \right) = g(x)\dfrac{d^2h(x)}{dx^2} + \dfrac{d g(x)}{dx}\dfrac{dh(x)}{dx}$
Can I then apply the derivative theorem to each operator, or does the product of gradients forbid this?
Does the fact that $g(x)$ is dependent on x forbid me from applying Fourier transforms?