In quantum mechanics, we work with linear operators on Hilbert spaces $\mathscr{H}$. Suppose I have two bounded ones, defined on the same space $A, B: \mathscr{H}\to\mathscr{H}$.
It seems to me there is an ambiguity on the way to deal with the derivative.
On one way, the operator $AB$ is usually interpreted as the composition $(A\circ B)f:=A(B(f))$ for every test function $f$ on $\mathscr{H}$. If so, the derivative $D$ operator on $AB$ should act as follows $$ D[AB]f = D[(A\circ B)f]= D[A(Bf)]D(B(f)) $$ On the other way, the following right examples treat $AB$ as if it were literally a product of operators instead of a composition. In other words, the preferred way to compute the derivative is the Leibniz rule $$ D[AB]= AD[B]+BD[A] $$
$1^{\rm{st}}$ example, by E. Pisanty:
The exponential of an operator $\hat A(t)$ does not obey the differential equation $$ \frac{d}{ dt}e^{\hat A(t)} \stackrel{?}{=} \frac{d \hat{A}}{ dt} e^{\hat A(t)} $$ that one might naively hope to satisfy. To see why this does not work, consider the series expansion of the exponential \begin{align*} \frac{d}{dt}e^{\hat A(t)} & = \frac{d}{dt}\sum_{n=0}^\infty \frac{1}{n!} = \sum_{n=0}^\infty \frac{1}{n!} \frac{d}{dt} \hat A^n(t), \end{align*} When we apply the product rule, we get the individual derivatives of each of the operators in the product, at their place within the product \begin{equation*} \frac{d}{dt} \hat A^n(t) = \frac{d\hat A}{dt} \hat A^{n-1}(t) +\hat A(t)\frac{d\hat A}{dt} \hat A^{n-2}(t) + \ \dots \ +\hat A^{n-2}(t)\frac{d\hat A}{dt} \hat A(t) +\hat A^{n-1}(t)\frac{d\hat A}{dt} \end{equation*} This can simplify to just $n\frac{ d\hat A}{dt} \hat A^{n-1}(t)$, in which case $\frac{d}{dt}e^{\hat A(t)} = \frac{d \hat A}{dt} \sum_{n=1}^\infty \frac{\hat A^{n-1}(t)}{(n-1)!} = \frac{d\hat A}{dt}e^{\hat{A}(t)}$, but only under the condition that $\hat A(t)$ commute with its derivative $$ \left[\frac{ d\hat A}{ dt} , \hat A(t)\right] \stackrel{?}{=} 0 $$
In this case, $A^n$ is seen as a product of $A$ with itself $n$ times, instead of $A \circ A \circ \dots \circ A$ $n$ times.
$2^{\rm{nd}}$ example, by Wikipedia:
The expectation value of an observable $A$, which is a Hermitian linear operator, for a given Schrödinger state $\vert\psi(t)\rangle$, is given by ${\displaystyle \langle A\rangle _{t}=\langle \psi (t)|A|\psi (t)\rangle .}$ In the Schrödinger picture, the state $\vert\psi(t)\rangle$ at time $t$ is related to the state $\vert\psi(0)\rangle$ at time $0$ by a unitary time-evolution operator $U(t)$: ${\displaystyle |\psi (t)\rangle =U(t)|\psi (0)\rangle .}$ In the Heisenberg picture, all state vectors are considered to remain constant at their initial values $\vert \psi(t)\rangle$, whereas operators evolve with time according to ${\displaystyle A(t):=U^{\dagger }(t)AU(t)\,.}$ The Schrödinger equation for the time-evolution operator is $${\displaystyle {\frac {d}{dt}}U(t)=-{\frac {iH}{\hbar }}U(t)}$$ where $H$ is the Hamiltonian and $\hbar$ is the reduced Planck constant. It now follows that $${\displaystyle {\begin{aligned}{\frac {d}{dt}}A(t)&={\frac {i}{\hbar }}U^{\dagger }(t)HAU(t)+U^{\dagger }(t)\left({\frac {\partial A}{\partial t}}\right)U(t)+{\frac {i}{\hbar }}U^{\dagger }(t)A(-H)U(t)\\&={\frac {i}{\hbar }}U^{\dagger }(t)HU(t)U^{\dagger }(t)AU(t)+U^{\dagger }(t)\left({\frac {\partial A}{\partial t}}\right)U(t)-{\frac {i}{\hbar }}U^{\dagger }(t)AU(t)U^{\dagger }(t)HU(t)\\&={\frac {i}{\hbar }}\left(H(t)A(t)-A(t)H(t)\right)+U^{\dagger }(t)\left({\frac {\partial A}{\partial t}}\right)U(t),\end{aligned}}}$$ where differentiation was carried out according to the product rule.
I really don't understand