The Formula to Calculate the Pressure of a Fluid Based on Depth
Formulas
The pressure of a fluid at a vertical distance $h$ below the surface, or simply, the fluid pressure at a depth of $h$, $P_{h}$, is as follows.
$$ P_{h}=P_{0}+\rho g h $$
Here, $P_{0}$ is the atmospheric pressure, $\rho$ is the density of the object, and $g$ is the gravitational acceleration.
Explanation
This formula applies only in static situations. It does not apply to moving fluids, i.e., fluids with velocity. Also, the shape of the container holding the fluid does not affect the application of this formula. That means, the pressure due to depth is independent of the container’s shape.
Derivation
Consider an object submerged at a depth of $h$ in a state of equilibrium. The pressure on each side of this object can be used to determine the fluid pressure at that depth. Let’s first identify all the forces acting on the object.
- Gravity$(\downarrow)$
- Force on the top surface$(\downarrow)$
- Force on the bottom surface$(\uparrow)$
- Force on the lateral surface$(\rightarrow \leftarrow)$
Let’s think about the horizontal direction first. The only force acting in the horizontal direction is the force on the lateral surfaces of the object. Therefore, since the object is in equilibrium, the force on one side is exactly opposite to the force on the other side. Thus, the sum of the horizontal pressures at depth $h$ is $0$. Although the figure shown is rectangular, this generally applies regardless of the shape of the object’s sides. 
Next, consider the vertical direction. Unlike the horizontal direction, in addition to the forces on the top and bottom surfaces, there is also the force of gravity. First, let’s define the direction of gravity as $+$. Since the mass of the object is volume$\times$density, and if we designate the density of the object as $\rho$, then the mass of the object is $m=\rho Ad$. Therefore, gravity is $mg=\rho Ad g$. The force on the top surface is $P_{h_{1}}A$ (force is pressure$\times$area). The force on the bottom surface is $-P_{h_{2}}A$. Since the object is in equilibrium, adding up all three forces gives us $0$.
$$ \rho Ad g + P_{h_{1}}A - P_{h_{2}}A=0 $$
By dividing the entire equation by $A$ and rearranging for $P_{h_{2}}$,
$$ P_{h_{2}}=\rho dg + P_{h_{1}} \tag{1} $$
At this point, substituting $h_{1}=0$ and $h_{2}=h$ from the figure above gives us $d=h_{2}-h_{1}=h$, and hence $(1)$ is
$$ P_{h}=\rho hg + P_{0} $$
$$ \implies P_{h}=P_{0} + \rho gh $$
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