$p \in [0,p_c]$.

Before defining invasion percolation, we record some basic properties of $g$.

\[pr:K\] The derivative from the right \[K-def\] $K := g'(p_c)$ exists and is given by $$\label{eq:K}
K := \frac{2}{p_c^3 \phi'' (1)} \, .$$

Let $\phi_p (z) := \phi (1 - p
+ p z)$ be the offspring generating function for the Galton-Watson tree thinned by $p$-percolation for $p \in (p_c,1)$. The fixed point of $\phi_p$ is $1 - g(p)$. In other words, $g(p)$ is the unique $s \in (0,1)$ for which $1 - \phi_p (1-s) = s$. The first two derivatives of $\phi_p$ at $1$ are given by $$\begin{aligned}
\phi_p'(1) & = & \frac{p}{p_c} \, ; \\[1ex]
\phi_p''(1) & = & p^2 \phi'' (1) \, . \\[1ex]\end{aligned}$$ By a Taylor expansion, this leads to $$1 - \phi_p (1-s) = \frac{p}{p_c} s - \frac{1+o(1)}{2}
   \phi'' (1) p^2 s^2$$ as $p \downarrow p_c$. Setting this equal to $s$ and solving for $s \in (0,1)$ yields the conclusion of the proposition. $\Cox$

\[cor:dersurvprob\]As $p \downarrow p_c$, $g'(p) \to K$.

The existence of $g'(p)$ on $(p_c, 1)$ follows from the implicit function theorem. To obtain an expression for $g'(p)$, we differentiate both sides of the expression $\phi(1 - p \cdot 