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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.

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The output of the differentiator is $Y(\omega)= j\omega X(\omega).$

Multiplying by the sine in the time domain is convolving with the Fourier Transform of sine in the frequency domain:

$$\mathcal F\left\{\sin(w_ct)\right\} = \frac{1}{2j}(\delta(\omega+\omega_c)-\delta(\omega-\omega_c))$$

Which results in the following expression for the output of the mixer, as a function of $Y(\omega)$:

$$\frac{1}{2j}(Y(\omega+\omega _c)- Y(\omega-\omega _c))$$

substituting back for $Y(\omega)$ in terms of $X(\omega)$:

$$\frac{1}{2j}(j(\omega+\omega _c) X(\omega+\omega _c)- j(\omega-\omega _c) X(\omega-\omega _c))$$

which is

$$\frac{1}{2}(\omega+\omega _c) X(\omega+\omega _c)- (\omega-\omega _c) X(\omega-\omega _c))$$

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