Geometric\Graphical verification of $ads=vdv$? In my Engineering Dynamics class, I've encountered this equation $ads=vdv$.  I could not wrap my head around why it was true, but I was able to come up with a verification of it as such: 
$\mathrm{ds}=s'(t)\mathrm{dt}=v\mathrm{dt}$ and $dv=v'(t)\mathrm{dt}=a\mathrm{dt}$.  This yields $a*v\mathrm{dt}=v*a\mathrm{dt}$, "proof" (such as it is) complete.  However, I still have some issues.  I assume if I have $a(b\mathrm{dt})$, it is equal to $b(a\mathrm{dt})$ because either they are constants or to be evaluated $t$ must be taken consistently, but I can't prove it formally.  
Also, I can't really see intuitively why this is true.  I can see why $\frac{dy}{dx}$ is the derivative, and graphically so.  Simply draw the graph where we take a value of $f'(x)$ and multiply it by $\Delta x$ to get $\Delta y$ (of course, $dy$ only equals an arbitrarily chosen $\Delta y$ if $f(x)$ is linear.)  But I can't even figure out how to derive the first relationship.
So I suppose the meat of my question is if someone would be so kind as to give a better (preferably graphical) description of why $ads=vdv$?
 A: i would say (if i understand correctly) your derived proof $a*v\mathrm{dt}=v*a\mathrm{dt}$ gives the desired understanding. 
The relation is about areas (if seen geometricaly), the area of the traversed space (times) the acceleration of traversal is equal to the area of velocity increase (times) velocity of traversal, symbolicaly:
$$ads=vdv$$
Why? Because acceleration is velocity increase and space traversed is (infinitesimaly) equal to velocity.
As a sidenote, Calculus (differential and integral), as used by Newton and Leibniz, actually stems from geometry and geometric considerations. The analytic treatment is a later development. They were able to produce valid and important results without analytic methods, exactly because they relied on geometry (see for example differential triangle by Barrow).
A: Before we can go into proving the formula $a\,ds=v\,dv$ we need a clear idea of what $ds$, $dv$, and $dt$ are. On a heuristic level they are "infinitesimal quantities" that are tied together in understandable ways. E.g., $ds=v\,dt$ is saying that the small increment of $s$ during the small time interval $dt$ is equal to the momentaneous velocity $v$ times $dt$. But note that the word "momentaneous" used here indicates that there is an additional variable lurking behind, namely $t$, which is not appearing in the formula $ds=v\,dt$.
For an arbitrary function $t\mapsto u(t)$ and a given point $t$ in time we have, by the definition of derivative,
$$u(t+T)-u(t)=u'(t)\, T \ +o\bigl(|T|\bigr)\qquad(T\to0)\ .$$
In the "modern" view of derivative this is expressed in "differential form" as
$$du(t):\quad  T\mapsto u'(t)\,T\tag{1}$$
whereby now the derivative of $u$ at time $t$ appears as a linear functional on the increment vectors $T$, small or large. For the "special" function $\bar t: \>t\mapsto t$ we of course have
$$\bar t(t+T)-\bar t(t)\equiv T\ ,$$
so that, trivially,  $$d\bar t(t): \quad T\mapsto T\ .$$
If we compare this with $(1)$ we can say that the functional $du(t)$ is a constant multiple of the functional $d\bar t(t)$, whereby the proportionality factor is $u'(t)$:
$$du(t)= u'(t)\>d\bar t(t)\ .$$
If we omit some $(t)$ and the bar over the $t$ in $(2)$ we arrive at the formula $du=u'(t)\>dt$,  which we are fond of since calculus 101.
After all this abstract nonsense the proof of the formula $a\,ds=v\,dv$ is as simple as you proposed: We have a base function $t\mapsto s(t)$, then the "derived" functions $v(t):=s'(t)$, and $a(t):=v'(t)$. Apply the principles explained here, and obtain
$$a(t) ds(t)= a(t) s'(t) dt =v(t)\>v'(t) dt= v(t)\>dv(t)\ .$$
This is not an equation about obscure infinitesimal quantities, but about linear functionals defined on the one-dimensional tangent space at $t\in{\mathbb R}$.
