Friedmann equation I've seen in literature
$$\dot{H} + H^2=\ldots$$
Source: https://en.wikipedia.org/wiki/Friedmann_equations
Defining the LHS. Since
$$H = \frac{\dot{a}}{a}$$
And that
$$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G}{3}(\rho + 3P)$$
Then replacing gives
$$H^2 = \frac{8\pi G}{3}(\rho + 3P)$$
So my question is how to you arrive at the additive Hubble term
$$\dot{H} + H^2 = \frac{8\pi G}{3}(\rho + 3P)?$$
 A: Note that $H=\frac{\dot{a}}{a}\implies\dot{H}+H^2=\frac{\ddot{a}}{a}$. You're asking about the special case $k=0,\,\Lambda=0$, but seem confused about what results we obtain. In this case, the Friedmann equations are$$H^2=\frac{\dot{a}^2}{a^2}=\frac{8\pi G\rho}{3},\,\dot{H}+H^2=\frac{\ddot{a}}{a}=-\frac{4\pi G(\rho+3p/c^2)}{3}.$$Hence$$\dot{H}=-4\pi G(\rho+p/c^2).$$
A: The forth equation you wrote, being
\begin{equation}
H^2=\frac{8\pi G}{3}(\rho+3P)
\end{equation}
is incorrect. The first Friedmann equation is given by
\begin{equation}
H^2=\frac{8\pi G}{3}\rho.
\end{equation}
Now, taking a time derivative in both sides, we find:
\begin{equation}
2H\dot{H}=\frac{8\pi G}{3}\dot{\rho}.
\end{equation}
Knowing that for a bariotropic fluid
\begin{equation}
\dot{\rho}=-3H(\rho+P),
\end{equation}
we have
\begin{equation}
\dot{H}=-4\pi G(\rho+P).
\end{equation}
Adding the first Friedmann equation to this one we obtain:
\begin{equation}
H^2+\dot{H}=-\frac{4 \pi G}{3}(\rho+3P),
\end{equation}
which is the correct form for the equation you are looking for (note that it agrees with the equations in the Wikipedia link you shared.
