Let be $f\in L^1(\mathbb{R})$,
I will be able to say that
$ \dfrac{\hat{df(w)}}{dx} = \int_{-\infty}^{\infty}\dfrac{df(x)}{dx}\exp(-2\pi j wx)dx $?
Why?
Let be $f\in L^1(\mathbb{R})$,
I will be able to say that
$ \dfrac{\hat{df(w)}}{dx} = \int_{-\infty}^{\infty}\dfrac{df(x)}{dx}\exp(-2\pi j wx)dx $?
Why?
When $f$ decays enough at $\pm\infty$ and $\hat{f}$ given by $\hat{f}(w)=\int_{-\infty}^{\infty}f(x)\exp(-ixw)\,dx,$ then \begin{eqnarray}\frac{d}{dw}\hat{f}(w)&=&\int_{-\infty}^{\infty}f(x)\frac{d}{dw}\exp(-ixw)\,dx\\ &=&-i\int_{-\infty}^{\infty}xf(x)\exp(-ixw)\,dx = -i\widehat{xf}(w) \end{eqnarray} (where in the last expression we abused the notation by writing $xf$ for the function $x\mapsto xf(x)$).
Your notation makes no sense - you cannot differentiate the FT with respect to $x$, as there is no dependence. (Well, OK, you get zero.) Here's how it works: consider
$\int_{-\infty}^{\infty} dx \, f'(x) e^{-i 2 \pi w x}$
(Sorry, I insist on using $i$ rather than $j$ - this is a math site after all).
Assume we can integrate by parts:
$\left [ f(x) e^{-i 2 \pi w x}\right]_{-\infty}^{\infty} + i 2 \pi w \int_{-\infty}^{\infty} dx \, f(x) e^{-i 2 \pi w x}$
In order for the above to make any sense,
$\lim_{x \to \infty} f(x) = 0$
Then we have that the FT of the derivative of $f$ is $i 2 \pi w \hat{f}(w)$. Differentiation in one domain becomes multiplication in another. Is this what you were after?