As seen in https://math.stackexchange.com/a/430885/634773, we can find the Fourier Transform of the derivative of a function through the anti-transform.
But if we do integration by parts, shouldn't it yield the same answer?
However, unlike what we see in https://math.stackexchange.com/a/430863/634773, the first term doesn't necessarily have a limit, or does it?
Sorry if I used bad english.
One useful set of assumptions is that $f,f'$ are square integrable on $\mathbb{R}$ and $f$ is absolutely continuous. But then their transforms must be interpreted as limits in $L^2$ of the truncated transforms. This simple set of assumptions works because $ff' \in L^1$, which forces $\int ff'dx = f^2/2$ to have limits at $\pm\infty$, and those limits must be $0$ in order for $f^2$ to integral. Then you can integrate by parts to legitimately obtain $\hat{f'}(\xi)=i\xi\hat{f}(\xi)$.