In my book "Quantum Theory for Mathematians" By B. Hall there is a discussion about the derivative of the inner product of a time-dependent wave functions $\psi(t)$ (note: no position dependence is indicated) (p. 72).
In the proof of a proposition in section 3.7 dealing with the time-dependent Schrödinger eqaution, stating $$\frac{d}{dt}\langle A\rangle_{\psi(t)} =\langle\frac{1}{i\hbar}[A,\hat{H}]\rangle_{\psi(t)}$$ where $\langle A\rangle_{\psi(t)}=\langle \psi(t), A\psi(t)\rangle$ and $A$ is an operator, there is a calculation showing that $$\frac{d}{dt}\langle\psi(t),A\psi(t)\rangle=\langle\frac{d\psi}{dt},A\psi\rangle+\langle\psi,A\frac{d\psi}{dt}\rangle.$$ Now, the inner product is defined to be the usual inner product for $L^2(\mathbb{R})$ with respect to position, but since $\psi$ is not dependent on position how can this inner product even be defined? And if the author meant for the wave function to be $\psi(t,x)$, should the (total) derivatives with respect to $t$ in the inner product not be partial derivatives with respect to $t$?
There are at least two plausible interpretations for the equation $$ \frac{d}{dt} \langle\psi(t), A\psi(t)\rangle = \left\langle\frac{d\psi}{dt}, A\psi\right \rangle + \left\langle\psi, A\frac{d\psi}{dt}\right\rangle; \tag{1} $$ not sure which, if either, the author intended.
The time-dependent wave function $\psi$ is taken to be $L^{2}$-valued, not complex-valued; the time derivatives are also $L^{2}$-valued, and (1) is a formal consequence of bilinearity of the inner product.
Equation (1) should be taken as holding at each position $x$, and the $L^{2}$ inner product found by integrating each side.
Technically, an element of $L^{2}$ isn't a complex-valued function, but an equivalence class of functions with any two identically equal except on a set of measure zero. Consequently, any equation involving pointwise (spatial) values of $\psi$ has to be interpreted carefully. It's possible this explains your author's notation.