Consider $f(A)=\rho^{\lvert A\rvert}$ where $0< \rho < 1$, $A\subset \mathbb Z^{d}$ and $\lvert \cdot \rvert$ the cardinality operator, I need to show that:
$\frac{d}{dt} E ^{A} f(A_{t}) \lvert _{t = 0}=[\rho^{-1}\lvert A\rvert-\lambda \lvert \{(x,y):x \in A, y\notin A, y \sim x\}\rvert](1-\rho)f(A)$
for the contact process on a tree of degree $d+1$, where $(A_{t}^{A})_{t \geq 0}$ denotes the graphical representation of the contact process, i.e.
$A_{t}^{A}:=\{y \in \mathbb Z^{d}: \text{there exists path from }(0,x) \text{ to } (t,y) \text{ for some }x \in A\}$
and $\lambda$ is the parameter of the contact process, i.e. the flip rates satisfy
$c(x,\eta)=1$ if $\eta(x)=1$ and $c(x,\eta)=\lambda\lvert \{y\sim x:\eta(y)=1\}\rvert$ if $\eta(x)=0$
My idea: Since we have the following relation for a generator $\mathcal{L}$,
$$ \frac{d}{dt}P_{t}f=P_{t}\mathcal{L}f=\mathcal{L}P_{t}f$$
and using $E^{A}f(A_{t})=P_{t}f(A)$, then we could state that
$$ \frac{d}{dt} E ^{A} f(A_{t}) \lvert _{t = 0}=\mathcal{L}f(A)$$
Some I am assuming I would need the generator of the contact process? If so, would the generator of the graphical representation be of the following form:
$\mathcal{L}f(A)=\sum\limits_{x \in \mathbb Z^{d}}q(A,A\cup{\{x\})[f(A\cup \{x\})-f(A)]}+\sum\limits_{x\sim y,\; y\notin A, \; x \in A}q(A,A\cup{\{y\})[f(A\cup \{y\})-f(A)]}+\sum\limits_{x \in A}q(A,A\setminus{\{x\})[f(A\setminus \{x\})-f(A)]}$