Maxwell's third equation is the differential form of Faraday's law induction.i.e
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$\overrightarrow{\nabla} \times \overrightarrow{E}=- \frac{\partial{\overrightarrow{B}}}{\partial{t}}$
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Derivation:
According to Faraday's Induced law-
$e=-\frac{\partial{\phi_{B}}}{\partial{t}} \qquad(1)$
According to Gauss's law of magnetism-
$\phi_{B}=\oint_{S} \overrightarrow{B}.\overrightarrow{dS} \qquad(2)$
Now substitute the value of $\phi_{B}$ in equation $(1)$
$e=-\frac{\partial}{\partial{t}} \oint_{S} \overrightarrow{B}.\overrightarrow{dS}$
$e=-\oint_{S} \frac{\partial{\overrightarrow{B}}}{\partial{t}}.\overrightarrow{dS} \qquad(3)$
The line integral of the electric field around a closed loop is called electromotive force. Thus
$e=\oint_{l} \overrightarrow{E}.\overrightarrow{dl} \qquad(4)$
from equation $(3)$ and $(4)$
$\oint_{l} \overrightarrow{E}.\overrightarrow{dl}=-\oint_{S} \frac{\partial{\overrightarrow{B}}}{\partial{t}}.\overrightarrow{dS} \qquad(5)$
According to Stroke's Theorem-
$\oint_{l} \overrightarrow{E}.\overrightarrow{dl}=\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{E}).\overrightarrow{dS} \qquad(6)$
from equation $(5)$ and equation $(6)$
$\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{E}).\overrightarrow{dS}=-\oint_{S} \frac{\partial{\overrightarrow{B}}}{\partial{t}}.\overrightarrow{dS}$
$\oint_{S} [(\overrightarrow{\nabla} \times \overrightarrow{E})+ \frac{\partial{\overrightarrow{B}}}{\partial{t}}].\overrightarrow{dS}=0$
If the surface is arbitrary then-
$(\overrightarrow{\nabla} \times \overrightarrow{E})+ \frac{\partial{\overrightarrow{B}}}{\partial{t}}=0$
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$\overrightarrow{\nabla} \times \overrightarrow{E}=- \frac{\partial{\overrightarrow{B}}}{\partial{t}}$
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This is Maxwell's third equation.
