The product rule works essentially as you think it might. You just must be careful about the commutation of of operators. For example, $$\frac{d}{dt}\big(AB \big) = \left(\frac{d}{dt}A\right) B + A \left(\frac{d}{dt}B\right) \neq \left(\frac{d}{dt}A\right) B + \left(\frac{d}{dt}B\right)A$$
If you would like a rigorous description of operator differentiation, we need to be more precise. Let $A:t \mapsto A(t)$ be a family of operators on some (for now, finite-dimensional) Hilbert space $\mathscr H$ which are indexed by a continuous variable $t$. The derivative $A'(t)$ is the operator such that, for arbitrary $\psi\in \mathscr H$,
$$A'(t) \psi = \lim_{\epsilon\rightarrow 0} \frac{A(t+\epsilon)\psi - A(t)\psi}{\epsilon}$$
assuming that this limit exists. Alternatively, $A'(t)$ is the operator such that $$\lim_{\epsilon_{\rightarrow 0}} \left\Vert \big[A(t+\epsilon)- A(t) - \epsilon A'(t)\big]\psi \right\Vert\rightarrow 0$$ for all $\psi\in \mathscr H$. Therefore, as a computational tool we may take an expression which depends on $t$, substitute $t\rightarrow t+\epsilon$, make first order replacements of the form $f(t+\epsilon) = f(t) + \epsilon f'(t)$, and then read off the term proportional to $\epsilon$ at the very end.
For example, $$A(t)B(t) \rightarrow A(t+\epsilon)B(t+\epsilon) \rightarrow \big(A(t) + \epsilon A'(t) \big) \big( B(t) + \epsilon B'(t)\big)$$ $$\rightarrow A(t) B(t) + \epsilon \underbrace{\bigg(A'(t) B(t) + A(t) B'(t)\bigg)}_{\frac{d}{dt} A(t) B(t)} + \epsilon^2 A'(t)B'(t)$$