Physical significance of divergence In my textbook 
They considered a parallelopiped $ABCDEFGH$ with sides $dx,dy,dz$ parallel to $x,y,z$ axis respectively 


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*$\vec V$ represents the vector velocity of the fluid at the centre $P$ of f the volume (parallelopiped) with components $V_x,V_y,V_z$ along three axis

*the fluid flows in or out through the all six faces.

*let us consider the fluid flow through the two opposite faces 1 and 2 of the volume element each being normal to x-axis and has area=dydz

*The value of the x component of $\vec V$ at the centre of the face 1 and 2 will be different from $V_x$ at the at the centre $P$.

*$\frac {\partial V_x}{\partial x}$ is the rate of change of $V_x$ along x-axis, then the change in value of $V_x$ in going from $P$ to the centre of vertical faces 1 or 2 = $\frac{\partial V_xdx}{\partial x2}$

*the value of x-component of the velocity $\vec V$ at the centre of face 
1 = $V_x-\frac{1}{2}\frac {\partial V_x}{x}dx$

Now my problem is


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*"The value of the $x$ component of $\vec V$ at the centre of the face 1 and 2 will be different from $V_x$ at the at the centre $P$" 
Why?

*$\frac {\partial V_x}{\partial x}$ is the rate of change of $V_x$ along $x$-axis, then the change in value of $V_x$ in going from $P$ to the centre of vertical faces 1 or 2 = $\frac{\partial V_xdx}{\partial x2}$

*$\frac {\partial V_x}{\partial x}$ is the rate of change of $V$ along x-axis because here water is flowing in all directions and we are concerned with the $x$ axis only, okay. But, change in the value of $V_x$ in going from $p$ to the centre of vertical faces of 1 or 2 = $$\frac{\partial V_xdx}{\partial x2}$$ but how and why. Please explain

*the value of $x$-component of the velocity $\vec V$ at the centre of face 
1 = $V_x-\frac{1}{2}\frac {\partial V_x}{x}dx$
How and why please explain.
 A: “The value of the  component of $\vec{V}$ at the centre of the face 1 and 2 will be different from $_x$ at the centre .”
“The $x$ component of $\vec{V}$” And “$V_x$” are the same thing. This just says $V_x$ is a function, so it’s value changes because $x$ changed going from center to a face. (See note about their bad terminology at the bottom of this.)

Unless otherwise specified, $\vec{V}$, and $V_x$ (the $x$-component of $V$), are specified at the point being considered, point P. Let’s call that point (a,b,c).
The center of face I is at ($a - \frac{dx}{2}$, b, c), and center of II is at ($a + \frac{dx}{2}$, b, c). If y =b and z=c don’t change, then on that line from center of I... to P... to center of II, everything is only a function of x.
Because dx is small, we can assume the change in $V_x$ is just the change in $x$ times $\frac{\partial V_x}{\partial x}$.  But notice that the change in $x$ is $\frac{d x}{2}$.
(That’s what $\frac{\partial V_x}{\partial x}$ means: how much $V_x$ changes per change in $x$)

The value of x-component of the velocity ⃗ at the centre of face 1 $= V_x- \left( \frac{\partial V_x}{\partial x}\right)~\left( \frac{dx}{2}\right)$ How? Why?
Change in $V_x$ going from P to face I
= Change in $V_x$ from $x=a$ to $x=a - \frac{dx}{2}$
= Change in $V_x$ from decreasing x by $ \frac{dx}{2}$
$=~ \left( \frac{\partial V_x}{\partial x}\right)~\left(\Delta x\right)=~ \left( \frac{\partial V_x}{\partial x}\right)~\left(- \frac{dx}{2}\right) $ $$\implies V_{x~\text{(face-}I)}= V_{x~(P)}- \left( \frac{\partial V_x}{\partial x}\right)~\left( \frac{dx}{2}\right)$$
Note, they seem at times to be implying “When I say $V_x$, Im talking about the x-component of $V$, at point P, but when I say ‘the $x$ component of V’ that could be referring to anywhere”. Which is dumb.
A: The velocity $V$ is actually a vector field i.e it has different values of velocity at different points in space. That is why you get different components at different points.
Rate of change of $x-component$ of velocity in the x- direction is $\frac{\partial V}{\partial x}$. Now since the centre of parallelopiped is $\frac{dx}{2}$ units away from the centres of faces of parallelopiped, we multiply by $\frac{dx}{2} $ to obtain change in velocity. 
Using the same logic, since the centre of face 1 is $\frac{dx}{2}$ away from the centre of parallelopiped, we subtract the change in velocity from the velocity at the centre.
