chain rule with position, velocity and acceleration function let assume i have a position function in 1 dimension with constant acceleration.
$$ x(t) = x_0 + v_0t + \frac{1}{2}at^2 $$
then it's first derivative is a velocity function:
$$ \frac{dx}{dt} = v(t) = v_0 + at $$
then it's second derivative is an acceleration function:
$$ \frac{dv}{dt} = a(t) = a $$
so in conclusion if we have x(t) a position function and we take a first derivative, we will get a velocity function and if we take it's second derivative we will get an acceleration function. this is what everyone knows.
now i see some lecture video says that:
$$ \frac{dv}{dt} = \frac{dv}{dx} * \frac{dx}{dt}  $$
if this is true then if i calculate $\frac{dv}{dx}$ and multiply by $\frac{dx}{dt}$ i will also get $a(t)$ but i don't know how to do it because when we apply chain rule we need to determine what is inner function and what is outer function but here there is only one function which is x(t) how to find $\frac{dv}{dx}$? 
can someone rewrite the position function into inner part and outer part?
or what is the valid way to do the calculation?
I'm very new to calculus and physic please explain step by step and easy simple example
 A: Note that these equations only hold for constant acceleration, so it's incorrect to write $a = a(t)$. 
You can isolate $t$ from the velocity equation
$$ t = \frac{v - v_0}{a} $$
Plugging into the displacement gives
$$ x = x_0 + \frac{v_0(v-v_0)}{a} + \frac{1}{2}\frac{(v-v_0)^2}{a} = x_0 + \frac{v^2 - {v_0}^2}{2a} $$
Solving for $v$ explicitly will yield an ambiguous square root unless we restrict $v > 0$, so instead we can employ implicit differentiation to get
$$ 1 = \frac{v}{a} \frac{dv}{dx} $$
which satisfies
$$ a = v\ \frac{dv}{dx} $$
A: $$ \frac{dv}{dx} = \frac{d(v_0+at)}{dx} = a\frac{dt}{dx} \tag{$\ast$}$$ Here, $v_0$ and $a$ is assumed to be purely independent constants. To find $\frac{dt}{dx}$, we can either write time in terms of position or differentiate the position function with respect to position which I prefer to do. So, we have $x = x_0+v_0t+\frac{1}{2}at^2$, differentiating both sides with respect to $x$, we get:
$$\frac{dx}{dx}=1=\frac{d(x_0+v_0t+\frac{1}{2}at^2)}{dx} = v_0\frac{dt}{dx}+\frac{1}{2}a(2t\frac{dt}{dx})=(v_0+at)\frac{dt}{dx} \\ \implies \frac{dt}{dx}=\frac{1}{v_0+at}$$ Substitute this into $(\ast)$ and multiply with $\frac{dx}{dt}=v(t)=v_0+at$ :
$$\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt}=(\frac{a}{v_0+at})(v_0+at)=a$$
