Would I be right in thinking that the operator $\hat O'(t)$ is different from the operator $D\hat O(t)$ where $D={d\over dt}$, since when acting on a function $f$, the second corresponds to $\hat O'(t)f+\hat O(t){d\over dt}f$?
Suppose $\hat O(t)$ is an operator acting on a space of functions $f=f(x,t)$.
I wish to find an antisymmetric operator $\hat P$ such that $\hat O'(t)=P\hat O$. As commented above, I don't think $P={d\over dt}$ would work. Can anyone help me out? Thanks.