Conjugate of an operator applied to a function In section 6.3.1 of the following MIT Open Course Ware (PDF) (22-02, Introduction to Applied Nuclear Engineering), the author, Prof Paola Cappellaro, derives Heisenberg's equation using the definition of expectation values,
$$
\frac{d}{dt}\langle\hat{A}\rangle=\frac{d}{dt}\int d^3x\psi^*(x,t)\hat{A}\psi(x,t)\\
=\int d^3x\left(\frac{\partial\psi^*}{\partial t}\hat{A}\psi+\psi^*\frac{\partial\hat{A}}{\partial t}\psi+\psi^*\hat{A}\frac{\partial\psi}{\partial t}\right)
$$
and Schrodinger's equation,
$$
\frac{\partial\psi}{\partial t}=-\frac{i}{\hbar}\hat{H}\psi,\qquad
\frac{\partial\psi^*}{\partial t}=\frac{i}{\hbar}\left(\hat{H}\psi\right)^*
$$
Then the author uses $\left(\hat{H}\psi\right)^*=\psi^*\hat{H}^*=\psi^*\hat{H}$ to get
$$
\frac{d}{dt}\langle\hat{A}\rangle=\int d^3x\left(\frac{i}{\hbar}\psi^*\left[\hat{H}\hat{A}-\hat{A}\hat{H}\right]\psi+\psi^*\frac{\partial\hat{A}}{\partial t}\psi\right)\\
=\frac{i}{\hbar}\langle\left[\hat{H},\,\hat{A}\right]\rangle+\langle\frac{\partial\hat{A}}{\partial t}\rangle
$$
There are two things I don't understand in this derivation. The first is that the total time derivative $\frac{d}{dt}$ turns into a partial time derivative $\frac{\partial}{\partial t}$ when inside the integral.
The second is that the author states
$$(\hat{H}\psi)^*=\psi^*\hat{H}$$
This confuses me as $\hat H$ is the Hamiltonian operator, and contains an $x$ derivative. Thus the left hand side of the above equation is another state $\phi^*$ such that $\phi=\hat H\psi$, but the right hand side is an operator: the derivative is yet to be applied to anything. This equation by itself just makes no sense to me.
Any help with either of these two issues would be greatly appreciated!
 A: Neither left-hand side nor right-hand side are properly operators that act on states and give states. They are a form of operators that act on states and give complex numbers. We call such a thing a bra (and a state is called a ket, so when you pair them you get a bra(c)ket). 
The left-hand side acts like $$((H\psi)^*)(\varphi) = \int \psi^* H \varphi \, dx.$$
If this confuses you, think about the finite-dimensional case. Then $\psi$ is a column vector and $H$ is a square matrix. While there is a complex conjugation operation for vectors, what you really want is the Hermitian conjugate, which is transpose and complex conjugation, denoted with a ${}^\dagger$. Then $(H\psi)^\dagger$ is the transpose of a column vector, so it's a row vector, and $(H\psi)^\dagger = \psi^\dagger H$ and this can act on a column vector with matrix multiplication.
For wavefunctions, $(\psi^\dagger)(x) = \psi(x)^*$ where the star is complex conjugation, and matrix multiplication is of a row vector and a column vector is integration.
A: 
There are two things I don't understand in this derivation. The first is that the total time derivative $\frac{d}{d t}$ turns into a partial time derivative $\frac{\partial}{\partial t}$ when inside the integral.

Generally speaking, this is implied by the Leibniz intregral rule, which is a particular case of differentiation under the integral sign. For a function $f(x,t)$, the rule implies:
$$ \frac{d}{dt} \left(\int_{x_0}^{x_1} f(x,t)dx \right) = \int_{x_0}^{x_1} \frac{\partial f(x,t)}{\partial t} dx$$
See the links for derivations, and at least note that if the limits of integration are themselves time dependent functions, their derivatives become important and you will have two additional terms outside the integral sign. The Leibniz rule typically refers to the case where the limits are constants and so their derivatives (and hence the two extra terms) are zero.

The second is that the author states: $(\hat{H}\psi)^* = \psi^*\hat{H}$. This confuses me as $\hat{H}$ is the Hamiltonian operator, and contains an $x$ derivative.

Going along with Robin Ekman's answer, you'll find that it's often more useful and illuminating to think about these expressions with your linear algebra goggles on -- that is, thinking about operators as self-adjoint matrices, rather than differential operators acting on functions in the Cartesian basis. The latter gets real ugly, real fast when you get into 3-dimensional QM. Dirac (bra-ket) notation will be your best friend, I promise.
But now that we've stressed that enough, I suppose an example might be helpful in thinking about linear differential operators. Consider the 1-D momentum operator, $\hat{p} = \frac{\hbar}{i} \frac{d}{dx}$. Let's now show that this operator is Hermitian:
$$<f|\hat{p}g> = \int_{-\infty}^{\infty} f^* \frac{\hbar}{i}\frac{dg}{dx}dx$$
Now, apply integration by parts. This gives:
$$\int_{-\infty}^{\infty} f^* \frac{\hbar}{i}\frac{dg}{dx}dx = \frac{\hbar}{i}f^* g |_{-\infty}^{\infty} + \int_{-\infty}^{\infty} \left(\frac{\hbar}{i}\frac{df}{dx}\right)^* g dx = <\hat{p}f|g>$$
Note that the first term must be zero, given that wavefunctions in QM are square integrable, or "normalizable," which requires that they go to zero at $\pm \infty$.
[Example source: Griffiths Introduction to QM, section 3.2.1]
