What is the adjoint of the operator? What is the adjoint of the operator $\frac{d}{dx}$in quantum mechanics? 
I think the answer should be $\frac{d}{dx}$ because it has no imaginary term.
 A: The answer is $- \frac{d}{dx}$. Of course there are subleties about the domain, but this is physics. To show the result you get the minus from.the partial integration.
A: In terms of wave functions, we have (for $p=-i\hbar\frac{d}{dx}$)
\begin{align}
\int_{-\infty}^\infty dx \phi(x)^* \left(\hat p \psi(x)\right)&=
\int_{-\infty}^\infty dx \phi(x)^* \left(-i\hbar\frac{d}{dx}\psi(x)\right)\, ,\\
&= -i\hbar\phi^*(x)\psi(x)\Bigl\vert_{-\infty}^\infty-(-i\hbar)
\int_{-\infty}^\infty dx \left(\frac{d}{dx}\phi(x)\right)^* \psi(x) 
\end{align}
using integration by parts.  It is an assumption that $\psi(\pm\infty)\to 0$ so this removes the boundary term and we're left with 
$$
\int_{-\infty}^\infty dx \phi(x)^* \left(\hat p \psi(x)\right)=
\int_{-\infty}^\infty dx \left(-i\hbar\frac{d}{dx}\phi(x)\right)^* \psi(x) 
=\int_{-\infty}^\infty dx \left(\hat p\phi(x)\right)^* \psi(x) \, .
$$
This shows that $\hat p=-i\hbar\frac{d}{dx}$, with the condition $\psi(\pm\infty)\to 0$, is hermitian (or its own adjoint).  There are subtleties tied technical issues but with "reasonable" functions this will work.
You can work out by yourself what happens for $\frac{d}{dx}$ given the discussion on $\hat p$.  Moreover, it is very good to reflect on the assumption $\psi(\pm\infty)\to 0$ and other technicalities in the context of for instance this discussion on the quantum infinite well.
