Through the differentiator, the frequency response will be $j \omega X(j\omega)$, but what about through a mixer with $\sin(\omega_ct)?$
Will it be $$\frac{1}{2j} j\omega\left[X(j(\omega-\omega_c))-X(j(\omega-\omega_c))\right]$$ $$\text{or}$$ $$\frac{1}{2j} j(\omega-\omega_c)\left[X(j(\omega-\omega_c))-X(j(\omega-\omega_c))\right]$$
I'm not sure if for a mixer you replace all $\omega = \omega-\omega_c$ or not.