Deriving formulas of distance, velocity and acceleration

First, we know the average velocity is
$\bar{v} = \frac{\Delta x}{\Delta t}$
Thus the instantaneous velocity is
$v=\lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t} = \frac{dx}{dt}$
If we calculate the derivative of the instantaneous velocity, we can get the acceleration. In other words, if we calculate the second dericaive of displacement, we also can get the acceleration.
$a=\frac{dv}{dt}=\frac{d^2x}{dt^2}$
Now we can use the formula(3) to get the formula of instantaneous velocity with component of $a$, $t$, and $v_0$.
$v = \int a dt = at+C$
The value of C in fact is $v_0$.

Let t=0,
$v = C$
Thus,
$v_0 = C$
Now we got the formula of instantaneous velocity :
$v = at+v_0$
The derivative of displacement is the instantaneous velocity, thus the indefinite integral of the instantaneous welocity is displacement.
$x = \int v dt = \int (at+v_0) = \frac{1}{2}at^2 + v_0t + B$
The calue of B in fact is $x_0$.

Let t=0
$x = B$
Thus,
$x_0 = B$
So we had the formula of displacement:
$x = \frac{1}{2}at^2 + v_0t + x_0$
From the formula(7), we got
$t = \frac{v-v_0}{a}$
Plugging formula(12) into the formula(11), we had:
$x=\frac{1}{2}a(\frac{v-v_0}{a})^2+v_0\frac{v-v_0}{a}+x_0$
Finally,
$2a(x-x_0) = v^2-v_0^2$