Q:

Derivating operator acting on ket

I'm deducing a formula, and I used the "product rule" $\frac{\partial}{\partial t}(A|\phi>)=(\frac{\partial A}{\partial t})|\phi>+A\frac{\partial}{\partial t}|\phi>$.
I'm actually getting the result I was looking for, however, I want to make sure this precedure is correct, as $A|\phi>$ is not actually a product, but a new ket constructed by applying the operator A on the old ket $|\phi>$.

A:

Yes.  Just think about what $A(t+dt)|\phi(t+dt)\rangle-A(t)|\phi(t)\rangle$ is to first order in $dt$ and it should be clear why.