How to extract IF (instantaneous frequency) from Discrete Hilbert Transform?
I have a nice solution for you that I'm always using whenever I need to compute the instantaneous frequency of a discrete analytical signal and which works much better then the common approach using $tan(.)$ where you need to introduce thresholds and fiddle around with adding/subtracting $2\pi$. In addition to that it is even faster.

Given a signal $x(t)$ and the corresponding analytical signal $$\xi(t) = x(t) + i\mathcal{H}x(t)$$ with $\mathcal{H}$ being the hilbert transform operator the instantaneous frequency $\omega(t)$ can be computed as followed. We have $\omega(t)=\dot{\varphi}(t)$ with $\varphi(t)=\Im log\xi(t)$ being the phase of the analytical signal $\xi$ ($\Im$ denotes the imaginary part). Combining these equations we get
$$
\omega(t)=\Im\frac{d}{dt}log\xi(t)=\Im\frac{\dot{\xi}(t)}{\xi(t)}
$$
This can also be seen be as follows:
$$
  \Im\frac{\dot{\xi}}{\xi} = \Im\frac{ \dot{A}e^{i\varphi} + iAe^{i\varphi}\dot{\varphi} }{Ae^{i\varphi}}
                       = \Im\Big[ \frac{\dot{A}}{A} + i\dot{\varphi} \Big]
                       = \dot{\varphi}
$$
Discretizing this equation you can now compute the discrete instantaneous frequency from the discrete analytical signal:
$$ \omega[n]\approx\Im\frac{\xi[n+1]-\xi[n]}{\xi[n]} $$
or second order:
$$ \omega[n]\approx\Im\frac{\xi[n+1]-\xi[n-1]}{2\xi[n]} $$

Of course this is not the exact solution as it is still requires a discrete differentiation of the signal. But this could be done using standard techniques.