Time derivative in Schrödinger equation In quantum mechanics, a system is descibed by an element $|\psi\rangle\in\mathcal{H}$, where $\mathcal{H}$ is a Hilbert space.
Then on $\mathcal{H}$ (or on a dense subspace of $\mathcal{H}$), we can define the Hamiltonian operator $\mathbf{H}:\mathcal{D}(\mathbf{H})\rightarrow \mathcal{H}$, where $\mathcal{D}(\mathbf{H})$ is a dense subspace of $\mathcal{H}$ (called the domain of definition of $\mathbf{H}$).
Now, at time $t_0$, we consider a system described by $|\psi(t_0)\rangle\in \mathcal{H}$, and this system evolves with the Schrödinger equation : $$\forall t\ge t_0,\quad \mathbf{H}|\psi(t)\rangle=i\hbar \frac{\text{d}|\psi(t)\rangle}{\text{d}t},$$
where $|\psi(t)\rangle$ is the state of the system at time $t$.
But what does $\frac{\text{d}|\psi(t)\rangle}{\text{d}t}$ mean ? Because if we take the definition of the derivative, then we would have :
$$\frac{\text{d}|\psi(t)\rangle}{\text{d}t}=\lim\limits_{h\rightarrow 0}\frac{|\psi(t+h)\rangle-|\psi(t)\rangle}{h}, $$ but how do you define this limit of functions on $\mathcal{H}$ ? More precisely, does the sequence $(|\psi(t)\rangle)_{t\ge t_0}$ evolves in the same Hilbert space $\mathcal{H}$ and what is the domain of definition of $t\mapsto |\psi(t)\rangle$ to take the 'derivative' ?
 A: If you want to be formal, the function $\psi : \mathbb{R}\to \mathcal{H}, t\mapsto \lvert\psi(t)\rangle$ needs to be understood as a function between Banach spaces (every Hilbert space is in particular a Banach space). The correct notion of derivative is then the Fréchet derivative.
Note that this vector-valued function is much easier to differentiate formally than an operator-valued one is in general, see this question and its answers for the rigorous definition of the derivatives of operator-valued functions.
A: A Hilbert space is, by definition, an inner product space (meaning that it is also a normed space) and it is Cauchy Complete. Completeness means that all the normal definitions of limiting procedures go through as usual (with no more than minor notational changes). 
A: There are several points: how do you define the derivative? No problem: if $\psi_1$ and $\psi_2$ are two functions in $\cal H$, there is by definition a scalar product $(\psi_1,\psi_2)$. If the 2 are functions on $\mathbb R$, the scalar product is often defined as
$$
(\psi_1,\psi_2)=\int_{-\infty}^\infty dx\,\psi_1^*(x)\psi_2(x)
$$
But a scalar product is the product of norms with the cosine of the angle, so the norm of $\psi_1$, its distance to the origin, is
$$
||\psi_1||^2=(\psi_1,\psi_1)
$$
With this norm you define a distance between two functions $\psi_1$ and $\psi_2$ as follows
$$
||\psi_1-\psi_2||=\left[
(\psi_1-\psi_2,\psi_1-\psi_2)
\right]^{1/2}
$$
A series of vectors $\psi_n$ tends to a limit $\psi_\infty$ if the distance $||\psi_n-\psi_\infty||\to0$ as $n\to\infty$.
Now the derivative $d/dt\left|\psi(t)\right\rangle$ is what the quotients $1/h\left[\left|\psi(t+h)\right\rangle-\left|\psi(t)\right\rangle\right]$ tend to, in the sense defined above.
Now if you are asking how to compute this: just take the old-fashioned partial derivative with respect to $t$: in every reasonable case, this will be the answer. If it is not, it would mean there is no answer.
An example of a case where things go wrong might be
$$
\psi(x,t)=\frac1{1+x^2}\exp\left[ie^x t
\right]
$$
The function is in $\cal H$, namely $L^2$, for all $t$, but the partial derivative with respect to $t$ definitely is not. This means, I think, that the derivative of $\psi(x,t)$ with respect to $t$ does not exist. 
A: The Schrodinger equation is just a differential equation. Its solution are just differentiable functions. So time derivation is a standard operation.
