Derivative of adjoint operator-valued function Consider an infinite dimensional complex Hilbert space $H$. I think that for a bounded operator-valued function $A: x\mapsto A(x) \in \mathcal B(H)$, where $x\in \mathbb R$, we can define the derivative $A^\prime(x)$ as the (unique) operator which obeys
$$\lim\limits_{h\to 0} \left\|A^\prime(x) - \tfrac{A(x+h)-A(x)}{h}\right\|_{\mathrm{op}} =0 \quad, \tag 1$$
if the limit exists. Here $\|\cdot\|_{\mathrm{op}}$ denotes the operator norm.
Now I think that from $\|A(x)^*\|_{\mathrm{op}}=\|A(x)\|_{\mathrm{op}}$, where $^*$ denotes the adjoint, we can show that the derivative of $A^*: x\mapsto A(x)^*$ exists at $x$ and is simply the adjoint of the derivative of $A$ at $x$, i.e we have
$$ (A^\prime(x))^* = (A^*)^\prime (x) \quad .\tag{2}$$
Question:
Can we find the same result as in $(2)$ if we define the derivative of bounded operator-valued functions in the strong operator topology instead of the uniform topology $(1)$? Or is there a weaker but similar result under some conditions?
 A: Here is a partial answer for an admittedly very restrictive special case.
To start, consider the following subspace:
$$\mathcal D(A^\prime(x)):=\left\{\psi\in H\,|\, \exists\, \psi_x \in H\, : \lim\limits_{h\to 0}\,\left\|\psi_x-\tfrac{A(x+h)-A(x)}{h}\psi \right\|=0\right\} \quad . $$
We find that for each $\psi \in \mathcal D(A^\prime(x))$, the corresponding $\psi_x$ is unique and this in turn allows us to define an operator $A^\prime(x): \mathcal D(A^\prime(x)) \longrightarrow H$ as $$A^\prime(x)\psi:=\psi_x$$
and extend by linearity. It is a well-defined linear operator and we call it the derivative of $A$ at $x$.$^1$
Now assume that $\mathcal D(A^\prime(x))=H$. I think that from the uniform boundedness principle it then follows that $A^\prime(x)$ is bounded$^2$, which in turn implies that $(A^\prime(x))^*$ is bounded and defined on the whole Hilbert space as well. With that we can show that $A^*: x\mapsto A(x)^*$ is differentiable in the weak sense and its derivative is $(A^\prime(x))^*$.
Indeed, by making use of the continuity of the underlying inner product $\langle \cdot,\cdot\rangle: H\times H \longrightarrow \mathbb C$, we find for all $\phi,\psi \in H$:
\begin{align}
\left\langle (A^\prime(x))^*\phi,\psi\right\rangle &= \langle \phi,A^\prime(x) \psi\rangle \\
&= \lim\limits_{h\to 0}\, \left\langle \phi,\tfrac{A(x+h)-A(x)}{h}\psi\right\rangle  \\
&= \lim\limits_{h\to 0}\, \left\langle \left(\tfrac{A(x+h)-A(x)}{h}\right)^* \phi,\psi\right\rangle \\
&= \lim\limits_{h\to 0}\left\langle \tfrac{A^*(x+h)-A^*(x)}{h}\phi,\psi\right\rangle \quad .
\end{align}

$^1$ I think this resembles the notion given here, p. 20-21.
$^2$ Thanks to @V. Moretti for pointing that out.
