Lets say that $\psi(t)$ maps from the time axis to a complex vector space, and we have an expression for $\partial_t\psi(t)$.
If there is an inner product $\langle\cdot{,}\cdot\rangle$ then we can use it to define a dual space (of covectors that map linearly from the vector space to the complex field). In particular, a time-dependent covector $I_\psi$ may be defined such that $I_\psi(v)=\langle\psi(t)\,{,}\,v\rangle$ for any vector $v$.
The time derivative of this covector shall be another linear map:
$$\begin{align}
(\partial_t I_\psi )(v)
&= \lim_{\epsilon\rightarrow0}
\frac{I_{\psi(t+\epsilon)}(v)-I_{\psi(t)}(v)}\epsilon
\\&= \lim_{\epsilon\rightarrow0}
\frac{\langle\psi(t+\epsilon)\,{,}\,v\rangle-\langle\psi(t)\,{,}\,v\rangle}\epsilon
\\&= \lim_{\epsilon\rightarrow0}
\frac{\big(\ \langle v\,{,}\,\psi(t+\epsilon)\rangle-\langle v\,{,}\,\psi(t)\rangle\ \big)^*}\epsilon
\\&=\lim_{\epsilon\rightarrow0}\
\big\langle v\,{,}\,\frac{\psi(t+\epsilon)-\psi(t)}\epsilon\big\rangle^*
\\&=\langle v\,{,}\,\partial_t\psi(t)\,\rangle^*
\\&=\langle\ \partial_t\psi(t)\ {,}\ v\,\rangle
\\&=I_{\partial_t\psi}(v)
\end{align}
$$
Now substituting in the Schrödinger equation,
$\partial_t\psi(t)=-\frac i \hbar H(\psi)$.
$I_{\partial_t\psi}(v)
=\langle-\frac i \hbar H(\psi)\ {,}\ v\,\rangle
=(-\frac i \hbar)^*\ \langle H(\psi)\ {,}\ v\,\rangle
=+\frac i \hbar\ \langle H(\psi)\ {,}\ v\,\rangle
= \frac i \hbar\ \langle \psi\ {,}\ H(v)\rangle
= (\frac i \hbar I_\psi H)(v)$
$\therefore \partial_t\,\langle \psi|=\frac i \hbar \langle\psi|H$
This result can be applied to finding the evolution of a density matrix operator
$\rho=|\psi\rangle\langle\psi|$.
$$
\begin{align}
\partial_t \rho &= \big(\partial_t|\psi\rangle\big)\langle\psi|+|\psi\rangle\big(\partial_t\langle\psi|\big)
\\&=\big(-\frac i \hbar H\ |\psi\rangle\big)\langle\psi|+|\psi\rangle\big(\frac i \hbar \ \langle\psi|\ H\big)
\\&=-\frac i \hbar\ H \rho +\frac i \hbar\ \rho H
\\\therefore i\hbar\ \partial_t\rho&=[H,\rho]
\end{align}
$$