Maxwell's fourth equation is the differential form of Ampere's circuital law.
$\overrightarrow{\nabla} \times \overrightarrow{H} = \overrightarrow{J} + \frac{\partial{\overrightarrow{D}}}{\partial{t}}$

Derivation:
According to Ampere's circuital law
$\oint_{l} \overrightarrow{B}. \overrightarrow{dl}=\mu_{0} i \qquad(1)$
According to Stroke's theorem
$\oint_{l} \overrightarrow{B}.\overrightarrow{dl}=\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{B}). \overrightarrow{dS} \qquad(2)$
From equation$(1)$ and equation$(2)$
$\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{B}). \overrightarrow{dS} = \mu_{0} i \qquad(3)$
where $i=\oint_{S} \overrightarrow{J}. \overrightarrow{dS} \qquad(4)$
So from equation$(3)$ and equation$(4)$
$\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{B}). \overrightarrow{dS} = \mu_{0} \oint_{S} \overrightarrow{J}. \overrightarrow{dS}$
$\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{B}). \overrightarrow{dS}  \mu_{0} \oint_{S} \overrightarrow{J}. \overrightarrow{dS}=0$
$\oint_{S} [(\overrightarrow{\nabla} \times \overrightarrow{B})  \mu_{0} \overrightarrow{J}]. \overrightarrow{dS}=0$
$(\overrightarrow{\nabla} \times \overrightarrow{B})  \mu_{0} \overrightarrow{J}=0$
$\overrightarrow{\nabla} \times \overrightarrow{B} = \mu_{0} \overrightarrow{J}$

We know that $\overrightarrow{B}=\mu_{0} \overrightarrow{H}$. So the above equation can be again written as
$\overrightarrow{\nabla} \times \overrightarrow{H} = \overrightarrow{J}$

Modified Maxwell's fourth equation:
The modified Maxwell's fourth equation is the differential form of the modified Ampere's circuital law.
We know the modified Ampere's circuital law
$\oint_{l} \overrightarrow{B}. \overrightarrow{dl}=\mu_{0} i + i_{d}$
Where $i_{d}$  Displacement current
So the derivation of the modified Maxwell equation is similar to the above derivation. Therefore the modified Maxwell's fourth equation can be written as
$\overrightarrow{\nabla} \times \overrightarrow{H} = \overrightarrow{J} + \overrightarrow{J_{d}} \qquad(1)$
Where $\overrightarrow{J_{d}}$  Displacement current density
And the value of $\overrightarrow{J_{d}}$ is
$\overrightarrow{J_{d}}=\epsilon_{0} \frac{\partial{\overrightarrow{E}}}{\partial{t}}$
$\overrightarrow{J_{d}}=\frac{\partial{\overrightarrow{D}}}{\partial{t}} \qquad(\because \overrightarrow{D}=\epsilon_{0} \overrightarrow{E})$
Now substitute the value of $\overrightarrow{J_{d}}$ in equation $(1)$, then
$\overrightarrow{\nabla} \times \overrightarrow{H} = \overrightarrow{J} + \frac{\partial{\overrightarrow{D}}}{\partial{t}}$

This is modified by Maxwell's fourth equation.