I'm reading through a derivation in a book and am having trouble understanding a step. Here's a screenshot
3.46 is the equation in $(k,\omega)$ space. They're doing an inverse Fourier transform back to $(k, t)$ space. I don't understand how they get the second term in 3.48. I'm pretty sure $\omega(k)$ are the roots of $D(k,\omega)$, and it seems like they're using the residue theorem, but how do they get the $\frac{\partial D(k,\omega)}{\partial \omega}$ in the denominator? I don't remember any derivatives appearing in the residue of a function.
Assume that $\frac{1}{D(k,\omega)(\omega-\omega_f)}$ has a simple pole at $\omega_k$. Then, using L'Hospital's Rule, we have
$$\begin{align} \text{Res}\left(\frac{1}{D(k,\omega)(\omega-\omega_f)},\omega=\omega(k)\right)&=\lim_{\omega \to \omega(k)}\left((\omega-\omega(k))\frac{1}{D(k,\omega)(\omega-\omega_f)}\right)\\\\ &=\frac{1}{\left. \frac{\partial D(k,\omega)}{\partial \omega}\right|_{\omega=\omega(k)}(\omega(k)-\omega_f)} \end{align}$$
since $D(k,\omega(k))=0$.