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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?

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    I think you want to differentiate w.r.to $w$.2013-06-25

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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 $x\mapsto xf(x)$).

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    I see what you mean. BTW it took me a minute to realize that your notation made sense.2013-06-25
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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?

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    I just noticed that this was posted in October 2012. Better late than never. Btw, combining our answers gives the duality differentiation and the multiplicative action of polynomials.2013-06-25