= Solution
The printed $\mathcal H$ is the <conformal Hubble parameter>, $\mathcal H=aH=\dot a$; $\mathcal H^{-1}$ is the <comoving Hubble radius>, rather than the physical radius $H^{-1}$. Since the derivative requested is with respect to <cosmic time>,
$$
\frac d{dt}\mathcal H^{-1}=-\frac{\ddot a}{\dot a^2}
=\frac{4\pi G a\rho}{3\dot a^2}(1+3\omega)
=\frac\Omega{2a}(1+3\omega).
$$
Thus the <cosmic-time derivative of the comoving Hubble radius> has
$$
\boxed{A=\frac\Omega{2a}>0\quad(a>0,\ \rho>0,\ H\ne0).}
$$
For an expanding universe, the same factor $1+3\omega$ determines whether a small departure from flatness grows and whether the <comoving Hubble radius> grows. Ordinary matter with $\omega>-1/3$ therefore has both signs associated with the conventional <flatness problem> and <horizon problem>. Scales whose physical wavelengths now exceed the <Hubble radius> were even farther outside it, in relative terms, earlier in such an era, rather than being brought inside for causal equilibration. Accelerated expansion with $\omega<-1/3$ reverses both signs.
There is a qualification to the word “always”: an actual <horizon problem> depends on the complete past history, not just this local sign. In a flat, constant-$\omega$ hot <Big Bang> model with $\omega>-1/3$, $a\propto(t-t_B)^p$ with $p=2/[3(1+\omega)]<1$, so the comoving <particle horizon> $\int_{t_B}^t dt'/a(t')$ is finite. The conventional causal problem then accompanies the flatness instability. An earlier accelerated era or a different past boundary can change the causal conclusion even when the present-era <comoving Hubble radius> is growing. Thus the requested correspondence is valid in that usual expanding hot <Big Bang> setting, not a universal logical equivalence between global horizons and local stability. At $\omega=-1/3$ both local effects are marginal.
Back to article page
