Is this an acceptable use of $a \,dx = v \,dv$? From the chain rule we have 
$$a\, dx = \frac{dv}{dt}\, dx = dv\frac{dx}{dt} = v \,dv$$
This was introduced in an early chapter of my physics textbook and I have been getting a lot of mileage out of it. For example, in part (b) of this problem

A small object is placed at the top of an incline that is essentially frictionless. The object slides down the incline onto a rough horizontal surface, where it stops in 5.0 s after traveling 60 m. (a) What is the speed of the object at the bottom of the incline and its acceleration along the horizontal surface? (b) What is the height of the incline?

you are supposed to use the work-energy theorem, like so:

Let $x$ denote the hypotenuse of the incline and $\theta$ its slope. Then $mg \sin\theta$ is the component of $F_g$ parallel to the surface. 
$$\begin{align}W_\mathrm{net} &= \Delta K \\
mg \,x \sin\theta& = \frac{1}{2}m(v_1^2 -0) \end{align}$$
Given $v_1 = 24$ from part (a), we can solve for the height $x \sin\theta = 29.36$ m.

Instead, I noticed that $a\,dx = v\,dv$ implies $\bar a \Delta x = \bar v \Delta v$. The parallel component of the acceleration vector is $g \sin \theta$, $\Delta x = x$, $\bar v = \frac{1}{2}(v_1 - 0) = \frac{v_1}{2}$, and $\Delta v = v_1$. Then we have
$$\begin{align} \bar a \Delta x &= \bar v \Delta v \\
g \,x \sin\theta & = \frac{v_1^2}{2} \end{align}$$
which gets you the same thing.


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*Are there situations in which an approach like this wouldn't work? (I realize that I assumed constant acceleration when calculating $\bar v$, but anything beyond that?) 

*Is it correct (from a physics standpoint) to say that $a\,dx = v\,dv$ implies $\bar a \Delta x = \bar v \Delta v$? I would prove this mathematically by integrating on both sides, noting that the average value of a function $f(x)$ on an interval $(x_0, x_1)$ is $$\bar f(x) = \frac{1}{x_1 - x_0}\int_{x_0}^{x_1} f(x) dx$$

*Does my work amount to a derivation of the work-energy theorem, or a lucky circumvention of it?

*Silly question: If a student did this on a test, would you give them credit?

 A: 
  
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*Are there situations in which an approach like this wouldn't work? (I realize that I assumed constant acceleration when calculating $\bar v$, but anything beyond that?) 
  

Once you start working in more than one dimension, things will get more complicated. In that case, you have 
$$\mathbf{a} \cdot d \mathbf{x} = \mathbf{v} \cdot d \mathbf{v}$$
and you'll need to keep track of the vector directions. Also, in some cases, the averages you define will not be convenient to compute (as discussed below).

  
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*Is it correct (from a physics standpoint) to say that $a\,dx = v\,dv$ implies $\bar a \Delta x = \bar v \Delta v$? I would prove this mathematically by integrating on both sides, noting that the average value of a function $f(x)$ on an interval $(x_0, x_1)$ is $$\bar f(x) = \frac{1}{x_1 - x_0}\int_{x_0}^{x_1} f(x) dx$$

Sure, but you need to be careful with how you define these averages. You've defined 
$$\bar{a} = \frac{1}{\Delta x} \int a \, dx, \quad \bar{v} = \frac{1}{\Delta v} \int v \, dv.$$
In other words your $\bar{a}$ is an average over changes in $x$, while your $\bar{v}$ is an average over changes in $v$. But you could also define other averages, like
$$\bar{a}' = \frac{1}{\Delta t} \int a \, dt, \quad \bar{a}'' = \frac{1}{\Delta v} \int a \, dv, \quad \bar{a}''' = \frac{1}{\Delta a} \int a \, da, \quad \ldots.$$
So if you forget what kinds of averages you're using, you'll get the wrong answer. You got lucky here because the acceleration was a constant, and any average of a constant is the same.

  
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*Does my work amount to a derivation of the work-energy theorem, or a lucky circumvention of it?

The only essential step you need to prove the work-energy theorem is the chain rule, as I explain in detail here. Using the chain rule to conclude $a \, dx = v \, dv$ and then integrating both sides is the whole derivation of the work-energy theorem. (Though some introductory textbooks somehow make it sound much more complicated...)

  
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*Silly question: If a student did this on a test, would you give them credit?
  

I would, but in practice it depends on how good your teacher is. 
A: To supplement the other answer: in the calculus of a single variable a differential is equivalent to a derivative, that is derivative $$\dot{x} = \frac{d x}{dt}$$ is literally a ratio of two differentials $dx$ and $dt$, which enables manipulations like
$$\dot{x} = dx\frac{1}{dt}, \frac{1}{\dot{x}} = \frac{dt}{dx},$$
etc.
This is not the case when one deals with functions of many variables, where a differential is
$$df(\mathbf{x}) = \sum_i\frac{\partial f(\mathbf{x})}{\partial x_i}dx_i,$$
and $\frac{\partial f(\mathbf{x})}{\partial x_i}$ is a symbolic notation for a partial derivative rather than a ratio of $\partial f(\mathbf{x})$ to $\partial x_i$ (neither of which is defined).
