Why do time derivatives change to '$j\omega$' when taking the time derivative of a phasor? For example, Faraday's law in the time domain is written as $$\nabla \times \vec{\mathbf{E}} = - \frac{\partial\vec{\mathbf{B}}}{\partial t}$$
When using phasor notation, Faraday's law is written as $$\nabla \times \vec{\mathbf{E}} = -j\omega \mu \vec{\mathbf{B}}$$
Why does the time derivative, $\frac{\partial}{\partial t}$ change to $j\omega$?
 A: If $f(t)$ is a sinusoidal function of time, i.e.
$$
f(t)=F_0\cos(\omega t +\varphi)=\Re\left\{F_0\mathrm e^{i(\omega t +\varphi)}\right\}=\Re\left\{\underbrace{F_0\mathrm e^{i\varphi}}_{F}\,\mathrm e^{i\omega t }\right\}=\Re\left\{F\,\mathrm e^{i\omega t}\right\}
$$
so we can represent the function $f(t)$ through the phasor $F\,\mathrm e^{i\omega t}$.
Taking the first derivative we have
$$
f'(t)=\frac{\mathrm d}{\mathrm dt}F_0\cos(\omega t +\varphi)=\frac{\mathrm d}{\mathrm dt}\Re\left\{F\,\mathrm e^{i\omega t}\right\}=\Re\left\{\frac{\mathrm d}{\mathrm dt}F\,\mathrm e^{i\omega t}\right\}=\Re\left\{i\omega F\,\mathrm e^{i\omega t}\right\}
$$
so we can represent the derivative $f'(t)$ through the phasor $i\omega F\,\mathrm e^{i\omega t}$.
Let be $\mathcal{\pmb E}(\pmb r,t)$ and $\mathcal{\pmb B}(\pmb r,t)$ sinusoidal fields:
$$
\begin{align}
\mathcal{\pmb B}(\pmb r,t)&=\pmb B_0(\pmb r)\cos(\omega t +\varphi)=\Re\left\{\pmb B_0(\pmb r)\mathrm e^{i(\omega t +\varphi)}\right\}=\Re\left\{\pmb B(\pmb r)\mathrm e^{i\omega t}\right\}\\
\mathcal{\pmb E}(\pmb r,t)&=\pmb E_0(\pmb r)\cos(\omega t +\vartheta)=\Re\left\{\pmb E_0(\pmb r)\mathrm e^{i(\omega t +\vartheta)}\right\}=\Re\left\{\pmb E(\pmb r)\mathrm e^{i\omega t}\right\}
\end{align}
$$
where  $\pmb B(\pmb r)=\pmb B_0(\pmb r)\mathrm e^{i\varphi}$ and $\pmb E(\pmb r)=\pmb E_0(\pmb r)\mathrm e^{i\vartheta}$.
We have 
$$
\nabla\times \mathcal{\pmb E}(\pmb r,t)=\nabla\times \Re\left\{\pmb E(\pmb r)\mathrm e^{i\omega t}\right\}=\Re\left\{\nabla\times \pmb E(\pmb r)\mathrm e^{i\omega t}\right\}
$$
and
$$
\frac{\partial}{\partial t}\mathcal{\pmb B}(\pmb r,t)=\frac{\partial}{\partial t}\Re\left\{\pmb B(\pmb r)\mathrm e^{i\omega t}\right\}=\Re\left\{\frac{\partial}{\partial t}\pmb B(\pmb r)\mathrm e^{i\omega t}\right\}=\Re\left\{i\omega \pmb B(\pmb r)\mathrm e^{i\omega t}\right\}
$$

Hence the Faraday's law 
  $$
\nabla\times \mathcal{\pmb E}(\pmb r,t)=-\frac{\partial \mathcal{\pmb B}(\pmb r,t)}{\partial t}
$$
  using phasors becomes
  $$
\nabla\times \pmb E(\pmb r)=-i\omega\pmb B(\pmb r)
$$

In general for a "suitable" function $f(t)$ we use the Fourier transform defined as
$$F(\omega) =  \int_{-\infty}^{\infty} f(t) \, \mathrm e^{-i \omega t} \mathrm dt$$
and the inverse Fourier transform is
$$f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) \, \mathrm e^{i \omega t} \mathrm d\omega$$
So we have 
$$f'(t) = \frac{\mathrm d}{\mathrm dt}\!\left( \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) \, \mathrm e^{i \omega t} \mathrm d\omega \right)= \frac{1}{2\pi} \int_{-\infty}^{\infty}   i \omega \, F(\omega) \,\mathrm  e^{i \omega t} \mathrm d\omega$$
Hence the Fourier transform of $f'(t)$ is $ i \omega \, F(\omega)$.
Let be $\mathcal{\pmb E}(\pmb r,t)$ and $\mathcal{\pmb B}(\pmb r,t)$ time varying fields satisfying
$$
\nabla\times \mathcal{\pmb E}(\pmb r,t)=-\frac{\partial \mathcal{\pmb B}(\pmb r,t)}{\partial t}
$$
Let's take the time Fourier transform 
$$
\begin{align}
\mathcal{F}\{\mathcal{\pmb B}(\pmb r,t)\}&=\pmb B(\pmb r,\omega)=\int_{-\infty}^{\infty}\mathcal{\pmb B}(\pmb r,t)\,\mathrm e^{-i\omega t}\,\mathrm d t\\
\mathcal{F}\{\mathcal{\pmb E}(\pmb r,t)\}&=\pmb E(\pmb r,\omega)=\int_{-\infty}^{\infty}\mathcal{\pmb E}(\pmb r,t)\,\mathrm e^{-i\omega t}\,\mathrm d t
\end{align}
$$
Taking the Fourier transform of Faraday's law we have
$$
\mathcal{F}\left\{\nabla\times \mathcal{\pmb E}(\pmb r,t)\right\}=\nabla\times \mathcal{F}\left\{\mathcal{\pmb E}(\pmb r,t)\right\}=\nabla\times {\pmb E}(\pmb r,\omega)
$$
and using the Fourier property for the derivatives
$$
\mathcal{F}\left\{-\frac{\partial \mathcal{\pmb B}(\pmb r,t)}{\partial t}\right\}=-i\omega{\pmb B}(\pmb r,\omega)
$$
So we find

$$
\nabla\times {\pmb E}(\pmb r,\omega)=-i\omega{\pmb B}(\pmb r,\omega)
$$

A: The $j\omega$ comes from the simple time derivative of a phasor.
Assume $\vec{\mathbf{B}}$ in the time domain is $$\vec{\mathbf{B}} = A\cos{\omega t - \phi}.$$
As a phasor, $\vec{\mathbf{B}}$ is noted as
$$\vec{\mathbf{B}} = Ae^{j\omega t - \phi} = Ae^{j\omega t}e^{-\phi}.$$
Taking the time derivative of the phasor $\vec{\mathbf{B}}$,
$$\frac{\partial\vec{\mathbf{B}}}{\partial t} = j\omega Ae^{j\omega t}e^{-\phi} = j\omega \vec{\mathbf{B}}.$$
