How does one properly define the derivative of one operator-valued function?

In Quantum Mechanics we usually consider operator-valued functions: these are functions that take in real numbers and gives back operators on the Hilbert space of the quantum system.

There are several examples of these. One of them is when we work with the Heisenberg picture, where we need to consider functions $\alpha : \mathbb{R}\to \mathcal{L}(\mathcal{H})$ such that $\alpha(t)$ is the operator at time $t$.

Another example is when we deal with exponentiation of operators, like when building the time evolution operator:

$$U(t,t_0)=\exp\left(-i\frac{H}{\hbar}(t-t_0)\right),$$

Here, the $\exp$ is usually understood as being defined via the eigenvalues of $H$.

The point is, the idea of a function $\alpha : \mathbb{R}\to \mathcal{L}(\mathcal{H})$ appears quite often in Quantum Mechanics, and sometimes one needs to differentiate these. In practice we do it formally, using all the properties we would expect, but I'm curious about how one would properly define this.

If we were dealing with bounded operators, then we could use the operator norm, which is available for this kind of operator, and define the derivative as we can usually we do when there's some norm around.

The point is that in Quantum Mechanics most of the time operators are unbounded.

So in the general case, how can one define the derivative of one operator-valued function?

If we restrict to closed operators (such as self-adjoint operators) acting on a Hilbert space $H$, then it is possible to define a metric. The set of closed operators becomes then a (non-complete) metric space $\mathcal{C}(H)$. Before discussing (briefly) what the metric is, let me remark that $\mathcal{C}(H)$ is not a linear space, for in general it is not possible to sum two closed unbounded operators. The distance between closed operators $T$ and $S$ is defined, roughly speaking, as the gap between the graphs $G(T)$ and $G(S)$. The graph of an operator is a closed linear manifold in $H\times H$ defined by $$G(T)=\{(\varphi,\psi) \in H\times H \;, \; \varphi\in D(T)\; , \; \psi=T\varphi \}\; .$$ For all the details of the definition, see e.g. Kato's book of 1966 on perturbation theory of linear operators.
In $\mathcal{C}(H)$, we have thus a notion of convergence $T_n\to T$. Convergence in this sense (called by Kato "generalized sense") extends roughly speaking the convergence in norm of bounded operators. If the resolvent set $\varrho(T)$ of $T$ is not empty, the generalized convergence is equivalent to convergence in the norm resolvent sense, i.e. it is equivalent to the convergence in norm of the resolvents (as bounded operators): $$T_n\to_\mathrm{gen} T \Leftrightarrow (T_n-z)^{-1}\to_{\mathrm{norm}} (T-z)^{-1}\; ,\; \forall z\in \varrho(T)\; .$$ More precisely, there exists an $n^*\in \mathbb{N}$ such that $z\in \varrho(T_n)$ for any $n\geq n^*$, and the convergence of resolvents holds. Of course convergence in the generalized sense is equivalent to convergence in norm if the operators are bounded.
Nevertheless one has still a problem in defining the derivative, since as I remarked before it is not in general possible to sum two closed operators and obtain another closed operators. It is possible to give abstract conditions on (densely defined) $T$ and $S$ for them to densely define a closed operator $T+S$, see this paper. However as you may notice, things are getting messier and messier. Anyways, let $T_0\in \mathcal{C}(H)$ be a fixed densely defined closed operator. We denote by $\mathcal{C}_{T_0}(H)$ the set $$\mathcal{C}_{T_0}(H)=\{T\in \mathcal{C}(H), T-T_0\in \mathcal{C}(H)\}\; .$$ Remark that $\mathcal{C}_{T_0}(H)$ may as well be empty. Nevertheless, let now $\alpha:\mathbb{R}\to \mathcal{C}_{T_0}(H)$ for some $T_0$, and $\alpha(x)=T_0$. Then the derivative $\alpha'(x)$ can be defined in the usual way since $h^{-1}\Bigl(\alpha(x+h)-\alpha(x)\Bigr)$ is a closed operator: $$\alpha'(x)=\lim_{h\to 0}h^{-1}\Bigl(\alpha(x+h)-\alpha(x)\Bigr)\; ;$$ where the limit is intended in the generalized sense (provided it exists). However, we are still not assured that the derivative makes sense in another point $x'\neq x$, if $\alpha(x')\neq T_0$!