Maxwell's first equation is the differential form of Gauss's law of electrostatics.i.e
$\overrightarrow{\nabla}.\overrightarrow{E}= \frac{\rho}{\epsilon_{0}} $

Derivation:
According to Gauss's law for electrostatic
$\oint_{s} \overrightarrow{E}.\overrightarrow{dS}=\frac{q}{\epsilon_{0}} \qquad(1)$
For continuous charge distribution inside the surface
$q=\oint_{v}\rho.dV$
Where
$\rho$→Charge density
dV→Small volume
Now substitute the value of $q$ in equation $(1)$ then
$\oint_{s}\overrightarrow{E}.\overrightarrow{dS}=\frac{1}{\epsilon_{0}} \oint_{v}\rho.dV \qquad(2)$
Now according to Gauss's divergence theorem
$\oint_{s} \overrightarrow{E}.\overrightarrow{dS}= \oint_{v} \overrightarrow{\nabla}.\overrightarrow{E} \qquad (3)$
From equation$(2)$ and equation$(3)$, we can write the above equation
$\oint_{v} \overrightarrow{\nabla}.\overrightarrow{E}= \frac{1}{\epsilon_{0}} \oint_{v}\rho.dV $
$\oint_{v} \overrightarrow{\nabla}.\overrightarrow{E} \frac{1}{\epsilon_{0}} \oint_{v}\rho.dV=0 $
$\oint_{v} (\overrightarrow{\nabla}.\overrightarrow{E} \frac{\rho}{\epsilon_{0}})dV=0 $
On solving the above equation
$\overrightarrow{\nabla}.\overrightarrow{E} \frac{\rho}{\epsilon_{0}}=0 $
$\overrightarrow{\nabla}.\overrightarrow{E}= \frac{\rho}{\epsilon_{0}} $

This is Maxwell's first equation.