Acceleration in terms of displacement I am having problems understanding the derivation of acceleration in terms of displacement. The first step is fine:
$$a(x) = \frac{\mathrm dv(x)}{\mathrm dt}
= \frac{\mathrm dv(x)}{\mathrm dx} \frac{\mathrm dx(t)}{\mathrm dt} = \frac{\mathrm dv(x)}{\mathrm dx} v(t)
\tag{1}$$
However, the second step causes me problems, because the use of $v$ is ambiguous:
$$\frac{\mathrm d}{\mathrm dx}\left(\frac{v^2}{2}\right)
= v \frac{\mathrm dv}{\mathrm dx} \implies a(x)
= \frac{\mathrm d}{\mathrm dx}\left(\frac{v^2}{2}\right)
\tag{2}$$
It is not stated whether they mean $v(t)$ or $v(x)$ and neither of the alternatives makes sense to me:
Alternative 1: $v = v(x)$
$$\frac{\mathrm d}{\mathrm dx}\left(\frac{v(x)^2}{2}\right)
= v(x) \frac{\mathrm dv(x)}{\mathrm dx},$$
which is not equal to what we derived in (1).
Alternative 2: $v = v(t)$
$$\frac{\mathrm d}{\mathrm dx}\left(\frac{v(t)^2}{2}\right)
= v(t) \frac{\mathrm dv(t)}{\mathrm dx},$$
which is not equal to what we derived in (1) either.
I have a few theories; firstly that my interpretation of what $v$ is in terms of in (1) might be incorrect. Secondly it might be that I apply the chain rule incorrectly somewhere. I would love for someone to clarify what is wrong – or explain what $v$ is in terms of when writing $a(x) = \frac{\mathrm d}{\mathrm dx}\left(\frac{v^2}{2}\right)$.
 A: I would not use $v(x)$ at all as it can be ambiguous. Take $x(t)=x_0\sin\left(\frac{t}{T}\right)$ with $x_0$ constant for example, then $v(t)=\dot{x}(t)=\frac{x_0}{T}\cos\left(\frac{t}{T}\right)$ and you have the relation $T^2v(t)^2+x(t)^2=1$. As you see, we can't determine the sign of $v(t)$ from this relation. Take $t_n=2\pi n T$, then $x(t_n)=x_0\sin(2\pi n)=0$ and $v(t_n)=\frac{x_0}{T}\cos(2\pi n)=(-1)^n\frac{x_0}{T}$, so the value of $x(t)$ is not enough information to determine $v(t)$.
Therefore assume, that $x(t)$ is invertible and write the inverse function as $t(x)$. Keep in mind, that $v(t)=v(t(x))$ and $v(x)=v(x(t))$, which you should add when deriving. So instead of:
$$\frac{\mathrm d v(x)}{\mathrm dt}\;\text{and}\;
\frac{\mathrm d v(t)}{\mathrm dx}$$
you should write:
$$\frac{\mathrm d v(x(t))}{\mathrm dt}\;\text{and}\;
\frac{\mathrm d v(t(x))}{\mathrm dx}.$$
This shows that the expression in your first and third equation ($\frac{\mathrm d v(x)}{\mathrm dx}v(t)$ and $v(x)\frac{\mathrm d v(x)}{\mathrm dx}$) are indeed the same.
You should also not denote both of the functions by the same symbol $v$, but $v_\mathrm{t}$ and $v_\mathrm{x}$ for example. They are not the same, since we have $v_\mathrm{t}=v_\mathrm{x}\circ x$ (or $v_\mathrm{t}(t)=v_\mathrm{x}(x(t))$) as well as $v_\mathrm{x}=v_\mathrm{t}\circ t$ (or $v_\mathrm{x}(t)=v_\mathrm{t}(t(x))$), which yield each other using $x\circ t=t\circ x=1$.
If you rewrite your calculations using the above notation, there won't be any confusion any more. You will get the following relations:
$$\frac{\mathrm d v_x(x)}{\mathrm dx}
=\frac{\mathrm d v_t(t(x))}{\mathrm dx}
=\frac{\mathrm d v_t}{\mathrm dt}(t(x))\cdot\frac{\mathrm d t(x)}{\mathrm dx}$$
$$a(t)=\frac{\mathrm d v_t(t)}{\mathrm dt}
=\frac{\mathrm d v_x(x(t))}{\mathrm dt}
=\frac{\mathrm d v_x}{\mathrm dx}(x(t))\cdot\frac{\mathrm d x(t)}{\mathrm dt}
=\frac{\mathrm d v_x}{\mathrm dx}(x(t))\cdot v_t(t)$$
From the latter equation, you also get your first equation (but pay attention to $v_t$ and $v_x$ being different functions!):
$$a(x)
=\frac{\mathrm d v_x}{\mathrm dx}(x)\cdot v_t(t(x))$$
A: the acceleration $~a(t)~$ is
$$a=\frac{d}{dt}\,v(t)$$
and the velocity v is
$$v=\frac{dx}{dt}$$
assume that $~v_x(t)=v_x(x(t))$
hence
$$a=\frac{d}{dt}\,v_x(t)=\frac{dv_x}{dx}\,\frac{dx}{dt}=\frac{dv_x}{dx}\,v$$
Example:
$$v_x=2\,x(t)^2\quad,x(t)=a\,t^2\quad\Rightarrow\\
a=\frac{dv_x}{dx}\,\frac{dx}{dt}=4\,x(t)\,2\,a\,t=8\,a^2\,t^3$$
now with
$$v_x=2\,x(t)^2=2\,a^2\,t^4\quad\Rightarrow\\
a=\frac{dv_x}{dt}=8\,a^2\,t^3$$
equal result !!!
