We have the PDE
$$\frac{\partial}{\partial t} f(t, x) + L f(t, x) = 0$$
where $f$ is a function and $L$ is an operator acting on $f$. We can write this as $$ (\partial_t + L)f = 0$$ Now my book says that we take the "Fourier transform on both sides to get"
$$\partial_t \mathcal{F}[f](t, \omega) + \mathcal{F}[Lf](t, \omega) = 0$$
I don't understand where this comes from. The second term on the left makes sense, since you just took the Fourier transform on it, but I don't understand why in the first term, you can throw the Fourier transform inside the derivative? Shouldn't it be outside?
Note that you only took the Fourier transform in the $x$ variable, so of course the Fourier transform commutes with the time derivative. That is, $$ \mathcal{F}_x[\partial_t f](t,\omega)=\int\limits_{-\infty}^\infty (\partial_t f(t,x))e^{-ix\omega}\, dx=\partial_t\int\limits_{-\infty}^\infty f(t,x)e^{-ix\omega}\, dx=\partial_t (\mathcal{F}_x [f](t,\omega)).$$