Let $d\in\mathbb N$ and $\lambda$ denote the Lebesgue measure on $\mathcal B(\mathbb R)$. Moreover, let $$\mathcal Ff:=\frac1{(2\pi)^d}\int e^{-{\rm i}\langle x,\;\cdot\;\rangle}f(x)\:\lambda^{\otimes d}({\rm d}x)$$ denote the Fourier transform and $$\mathcal F^{-1}f:=\int e^{{\rm i}\langle x,\;\cdot\;\rangle}f(x)\:\lambda^{\otimes d}({\rm d}x)$$ denote the inverse Fourier transform of $f\in\mathcal L^1(\lambda^{\otimes d})$.
Let $\alpha\in\mathbb N_0^d$. By partial integration, we obtain that $$\int g\partial^\alpha f\:{\rm d}\lambda^{\otimes d}=(-1)^{|\alpha|}\int f\partial^\alpha g\:{\rm d}\lambda^{\otimes d}\tag1$$ for all $f\in C_c^\infty(\mathbb R^d;\mathbb C)$ and $g\in C^\infty(\mathbb R^d;\mathbb C)$.
If we fix $y\in\mathbb R^d$ and choose $g:=e^{{\rm i}\langle x,\;\cdot\;\rangle}$, we have $$\partial^\alpha g={\rm i}^{|\alpha|}y^\alpha g\tag2$$ and hence should obtain $$\mathcal F^{-1}(\partial^\alpha f)(y)=\underbrace{(-1)^{|\alpha|}{\rm i}^{|\alpha|}}_{=\:(-{\rm i})^{|\alpha|}}(\mathcal F^{-1}f)(y)\tag3$$ from $(1)$.
Question 1: However, in the Wikipedia article, they give the Formula $$\mathcal F(\partial^\alpha f)(y)={\rm i}^{|\alpha|}y^\alpha(\mathcal F)(y)\tag4,$$ which is of from $(3)$ by a factor of $(-1)^{|\alpha|}$, instead. I know that there is no standard way to define the (inverse) Fourier transform, but I'm still unsure whether I've made any mistake in my derivation of $(3)$. So, is $(3)$ correct?
Question 2: How can we extend $(3)$ to $f\in\mathcal S(\mathbb R;\mathbb C)$? I guess we need a suitable density argument ...
Question 1
Note that the formula in (3) is for the inverse Fourier transform while the formula in (4) is for the Fourier transform.
Question 2
Instead of $g\in C^\infty(\mathbb R^d;\mathbb C),$ take $g\in BC^\infty(\mathbb R^d;\mathbb C),$ the space of bounded smooth functions. Then it's enough that $\partial^\alpha f(x)\to 0$ as $|x|\to\infty.$