Is $f''(f)df=f'(x)df'$ correct?

This question is about kinematics and $a , v, x$ stand respectively for acceleration, velocity, position. Supposing we have an expression of $a$ in function of $x$, we have the following theorem:

$$v^2=v_0 ^2+2\int _{x_0} ^x a(x) \, dx$$

One proof I often saw is the following:

$$\int _{x_0}^{x}a \,dx=\int _{t_0}^{t}\dfrac {dv}{dt} \dfrac {dx}{dt}dt=\int _{v_0} ^{v}v\,dv=\frac{1}{2}v^2-\frac {1}2v_0 ^2$$

Now, i was trying to formulate the same proof in more rigorous terms. Putting $x=f(t)$ we have:

$$\int _{f(t_0)} ^{f(t)} f''(f)df= \int _{t_0} ^{t} f''(f(t))f'(t)dt$$

I don't know how to continue: I see that the result would follow if $$f''(f(t))dt=df' \qquad \iff \qquad f''(f(t))=f''(t)$$

but I don't think this is correct. Any help is appreciated.

EDIT - Answer: I think i've spotted the flaw of my reasoning, suggested by the comments above. Giving a function $a(x)$ means that we know a function of position which satifisfies the differential equation $$x''=a(x)$$. So, it's ok to substitute in the (first) integral $a(x)$ with $dv/dt=d^2x/dt^2$. What makes no sense (and that i didn't initially realize) is to write $f''(f(t))$ which is just symbolism. So (as one would expect) books' proofs are correct.

-
$v$ is velocity, not speed. Basically, speed is equal to $|v|$. – Thomas Andrews Feb 27 '13 at 12:44
No, that's not right, since $f''(f(t)) = f''(t)$ is quite illogical. – Cheeku Feb 27 '13 at 12:46
$f''$ takes a time as its argument, and $f$ is a distance, so the units of $f''(f)$ do not match up. It could be $a(f)$ since $a$ is a function of $x=f(t)$. – Thomas Andrews Feb 27 '13 at 12:49
@ThomasAndrews He can actually take $f''(f)$, but on the right side he uses replaces $\frac{dv}{dt}$ with $f''(f(t))$ which makes no sense to me. Changing that solves the problem. – Cheeku Feb 27 '13 at 12:54
@Cheeku I don't see how you an think that $f''(f)$ makes sense, since $f$ takes $t$ as an argument, which has time units, and $f$ returns position/distance units, so $f''(f)$ can't make sense. – Thomas Andrews Feb 27 '13 at 13:08

You can't do $f''(f)$ because $f''$ takes a time as an argument, and $f''$ is an acceleration.

If $f(t)$ describes a traversal from $x_0$ to $x_1$ in this condition, then the acceleration condition is that $f''(t) = a(f(t))$.

This means that the acceleration depends only on location - for example, an item in a gravity field, an item tied to a spring, the left-right position of a pendulum, or a charged particle in an unchanging magnetic field.

So, assume $f(t_0)=x_0$ and $f(t_1)=x_1$ and $f''(t)=a(f(t))$ for all $t\in[t_0,t_1]$.

By setting $x=f(t)$ we get: \begin{align}\int_{x_0}^{x_1} a(x) dx &= \int_{t_0}^{t_1} a(f(t))f'(t)dt\\ &=\int_{t_0}^{t_1} f''(t)f'(t)dt = \frac{1}{2}(f'(t))^2\bigg|_{t_0}^{t_1} = \frac{1}{2}(v_1^2-v_0^2) \end{align}

Note

There is an abuse of notation in the expression: $$\int_{x_0}^x a(x)dx$$ The variable $x$ should not be both a bound of the integral and the variable of integration, because otherwise inside the integral we are never sure which $x$ where are talking about.

-