Two different versions of Schrödinger's equation - are they equivalent? For simplicity, let's look at the case of one particle in one dimension. We usually think of the wave function as a function
\begin{align}
\Psi\colon\mathbb R\times[0,\infty[&\to\mathbb C\\
(x,t)&\mapsto\Psi(x,t)
\end{align}
and Schrödinger's equation has the form
\begin{equation}\tag{1}
\mathrm{i}\hbar\frac{\partial}{\partial t}\Psi=\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x{}^2}+V\right)\Psi\quad
\end{equation}
On the other hand, currying allows us to think of the wave function as a function of time:
\begin{align}
\psi\colon[0,\infty[&\to L^2(\mathbb{R},\mathbb{C})
\end{align}
In other words, $\psi(t)$ is the function defined by $(\psi(t))(x)=\Psi(x,t)$ - the particle's wave function at the time $t$.
That being said, the following equation makes sense:
\begin{equation}\tag{2}
\mathrm{i}\hbar\frac{\mathrm{d}}{\mathrm{d}t}\psi=\left(-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x{}^2}+V\right)\psi
\end{equation}
Please notice that I replaced the partial derivatives by total derivatives.
Q: Are both equations - (1) and (2) - valid? Are they equivalent?
 A: The two approaches are different and inequivalent. The physically correct one is the latter since the Schroedinger equation is nothing but the application of the Stone theorem to the time evolution and there the relevant topology to compute derivatives is the Hilbert space one. In other words, the S. equation arises by taking a suitable  type of $t$ derivative on both sides of the identity
$$\psi_t = e^{-itH/\hbar} \psi_0\:.$$
This type of derivative is the one described below in (1) which is used in the said theorem eventually leading to the wanted equation
$$i\hbar \dot{\psi}_t =  H \psi_t\:.$$
The other notion of $t$-derivative, relying on standard calculus only,  is however interesting from the viewpoint of PDE theory.
The correct notion of $t$-derivative in QM  is therefore  the one, I indicate by $\dot{\psi}_t$,  in the $L^2$ space performed with respect to the norm of that space:
$$\left|\left|\dot{\psi}_t - \frac{\psi_{t+h}-\psi_t}{h}\right|\right|_{L^2(\mathbb{R})} \to 0 \quad \mbox{for}\quad h\to 0\:.\tag{1}$$
In integral terms
$$\int_{\mathbb{R}}\left|\dot{\psi}_t(x) - \frac{\psi_{t+h}(x)-\psi_t(x)}{h} \right|^2 dx \to 0 \quad \mbox{for}\quad h\to 0\:.$$
The existence of the standard $t$-derivative $\partial_t \psi_t(x)$ is a $x$-pointwise condition, whereas the derivative $\dot{\psi}_t$ in the sense of $L^2$ space is $x$-global (integral) condition.They are quite independent from each other. The former does not have global implications whereas the latter sees the functions up to zero $x$-measure sets.
However if $t \mapsto \psi_t$ admits $L^2(\mathbb{R})$ $t$-derivative $\dot{\psi}_t$ and, simultaneously, $(t,x) \mapsto \psi_t(x)$ also admits standard partial $t$ derivative $\partial_t\psi_t(x)$, then the two derivatives coincide for almost  all $x\in \mathbb{R}$ at the given $t$.
PROOF Suppose that $\partial_t\psi_t(x)$ exists for all $x\in \mathbb{R}$ and the said $t$. Suppose that also
$\frac{d}{dt}_{L^2}\psi_t =: \dot{\psi}_t \in L^2(\mathbb{R})$ exists as in (1). Hence, if $h\to 0$,
$$\frac{\psi_{t+h}-\psi_t}{h} \to \dot{\psi}_t \quad \mbox{in $L^2(\mathbb{R})$ sense as in (1)}\:.$$
Due to a known theorem of $L^p$ convergence, there is a sequence $h_n \to 0$ such that, as $n\to +\infty$,
$$\frac{\psi_{t+h_n}(x)-\psi_t(x)}{h_n} \to \dot{\psi}_t(x) \quad \mbox{for almost all $x \in \mathbb{R}$}\:.$$
However the existence of $\partial_t\psi_t(x)$ requires
$$\frac{\psi_{t+h_n}(x)-\psi_t(x)}{h_n} \to \partial_t\psi_t(x) \quad \mbox{for  all $x \in \mathbb{R}$}\:.$$
In summary
$$\partial_t\psi_t(x) = \dot{\psi}_t(x) \quad \mbox{for almost all $x \in \mathbb{R}$.}$$
