Does $\frac{d}{dx}\langle u|= (\frac{d}{dx}|u\rangle)^\dagger$? Here $|u\rangle$ is any vector in a Hilbert space and $\dagger$ is the adjoint/conjugate-transpose.

It seems to me that this should not be the case since $(\frac{d}{dx}|u\rangle)^\dagger = \langle u| (\frac{d}{dx})^\dagger = - \frac{d}{dx} \langle u|$. Here we used that $(\frac{d}{dx})^\dagger=-\frac{d}{dx}$ since the momentum operator $p=-i \frac{d}{dx}$ is Hermitian so $p^\dagger = p$ and $p^\dagger = i (\frac{d}{dx})^\dagger$.

Related to Is the time derivative of the adjoint equal to the adjoint of the time derivative? , but possibly with a different conclusion for the space derivative and time derivative.

Does my logic make sense for why the derivative of the adjoint is not the adjoint of the derivative, or where did I make a mistake? Thanks!