# ∂p / ∂s ds Component in Euler's Fluids Equation

For the last hour I have been trying to understand what $\dfrac{\partial p}{\partial s}ds$ is in Euler's equation but I have a problem. You can see this image:  I know that $p\ dA = \text{Force}$

$γ ds dA = mg$

but what is $\left(p + \dfrac{\partial p}{\partial s} dA\right)$?

$\frac{\partial p}{\partial s}$ is the pressure gradient through the control volume. Multiplying it by $ds$, the length of the control volume, gives the change in pressure from one side of the control volume to the other. It is essentially a first-order Taylor-series expansion of the function $p(s)$. This method is used a lot in fluid dynamics derivations; the most thorough explanation I've seen is in "Computational Fluid Dynamics" by John Anderson.
Whenever a fluid flows in a pipe, it flows due to difference in pressure. Due to this difference in pressure, there is a pressure gradient which is $\dfrac{\partial p}{\partial s}$. $\dfrac{\partial p}{\partial s}\ \text{d}s$ is the change in pressure over the length $\text{d}s$. $p+\dfrac{\partial p}{\partial s}\ \text{d}s$ is the pressure at the exit in terms of pressure function $p$ at inlet + change in pressure over length $\text{d}s$. So, $\left(p+\dfrac{\partial p}{\partial s}*ds\right)dA$ is also the force like $p\cdot \text{d}A$ with pressure $p$ at the exit equals $p+\dfrac{\partial p}{\partial s}\ \text{d}s$.Pressure variation in fluid is given by Tailor's series expansion with respect to length $s$ about $p$. $p(s + \Delta s) = p(s) + \dfrac{\partial p}{\partial s}\ \Delta s + O(\Delta s2)$. $O(\Delta s2)$ is the higher order term which is neglected.