Suppose that the operator family $(A(t))_{t \in \mathbb{R}}$ with dense domain of definition $D(A(t)) = D \subseteq X$ is uniformly sectorial with angle $\phi < \pi/2$.
Additionally assume that their resolvents $R(\lambda,A(t)) = (\lambda - A(t))^{-1}$ commute, i.e. for $t,s \in \mathbb{R}$ and $\lambda, \mu$ in the sector $\Sigma_\phi$ we have that $R(\lambda, A(t))R(\mu, A(s)) = R(\mu,A(s)) R(\lambda,A(t))$.
If for all $x \in D$ the mapping $t \mapsto A(t)x$ is differentiable, can we also (strongly) differentiate the mapping $t \mapsto R(\lambda, A(t))x$? Maybe further assumptions are needed?
What would the derivative look like? ( a formal calculation suggests that $\frac{d}{dt} R(\lambda, A(t)) = A'(t) R(\lambda, A(t))^2$ )
Edit: I think this could work:
We use the second resolvent identity for $h > 0$ and all $t \in \mathbb{R}, \lambda \in \Sigma_\phi$: \begin{align*} R(\lambda,A(t + h)) - R(\lambda,A(t)) &= R(\lambda,A(t + h)) \; (A(t + h) - A(t)) \; R(\lambda,A(t)) \end{align*} In particular, we can see that $R(\lambda,A(t + h)) \to R(\lambda,A(t))$ strongly as $h \to 0$ by the uniform sectoriality and the strong continuity of the family $(A(t))$.
Furthermore, we have \begin{align*} \frac{1}{h}(R(\lambda,A(t + h)) - R(\lambda,A(t))) &= \; R(\lambda,A(t + h)) R(\lambda,A(t)) \; \frac{1}{h} (A(t + h) - A(t)) \end{align*} where we use the commuting resolvents to rearrange the terms.
Taking the limit the uniform boundedness of the resolvents implies \begin{align*} \frac{d}{dt} R(\lambda,A(t)) &= R(\lambda,A(t)) R(\lambda,A(t)) A'(t)\\ &= R(\lambda,A(t))^2 A'(t)\\ &= - \frac{d}{d\lambda} R(\lambda,A(t)) A'(t) \end{align*} where I used that $t \mapsto A(t)$ is strongly continuous and commutes with its resolvent on $D$.
Did I miss something?
