For each $x \in [a,b]$ let $A_x: H \to H$ be an operator on a Hilbert space.
The inner product $(A_xu,v)_{H}$ can be thought of as a function from $[a,b] \to \mathbb{R}.$ I want to say that $(A_xu,v)_{H}$ is differentiable in the classical sense $$\lim_{h \to 0}\frac{(A_{x+h}u,v)_{H}-(A_xu,v)_{H}}{h}=\lim_{h \to 0}\frac{(A_{x+h}u-A_xu,v)_{H}}{h}=(\lim_{h \to 0}\frac{A_{x+h}u-A_xu}{h},v)_{H}=0$$
Does it make sense to write the above? In what sense then is the limit on the RHS (the one inside the inner product) taken??
Here is an example to show that the existence of $$ \lim_{h \to 0}\frac{(A_{x+h}u - A_xu,v)_{H}}{h}$$ for all $u,v$ does not imply the existence of $$\lim_{h \to 0}\frac{A_{x+h}u-A_xu}{h}$$
The space $H$ will be $L^2[-1,1]$. Let $A_x=0$ for $x\le 0$, and $A_x f(t)=x \exp ( it/{x}) f(t)$ for $x>0$. For any $u,v \in H$ we have $$ \lim_{h \to 0}\frac{(A_{ h}u - A_0 u,v)_{H}}{h} = \lim_{h \to 0} \int_{-1}^1 \exp ( it/{h}) u(t)v(t)\,dt =0 $$ by the Riemann-Lebesgue lemma. On the other hand, $$ \frac{ A_{ h}u-A_0 u }{h }(t) = \exp ( it/{h}) u(t) $$ is a unimodular multiple of $u$, so it does not converge to $0$ (nor to anything else) as $h\to 0$, unless $u=0$.
Recommended reading: Weak operator topology.