We define an operator $$ Rv(x)=F^{-1}(1+\xi^{2})^{-\frac{1}{2}}Fv, $$ where $(Fv)(\xi)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}v(x)e^{-ix\xi}\ dx$.
Is this operator linear under derivatives? I mean, $$ \frac{\partial}{\partial x}Rv=R(\frac{\partial v}{\partial x}). $$
Note that $F\frac{\partial}{\partial x}v=i\xi Fv$, $\frac{\partial}{\partial x}F^{-1}v(\xi)=F^{-1}i\xi v(\xi)$. So, $$ R(\frac{\partial v}{\partial x})=F^{-1}\frac{i\xi}{(1+\xi^{2})^{\frac{1}{2}}}Fv=\frac{\partial}{\partial x}F^{-1}\frac{i\xi}{(1+\xi^{2})^{\frac{1}{2}}}Fv=\frac{\partial}{\partial x}Rv(x). $$
Am I right?