Valid justification for algebraic manipulation of $\mathrm{d}y/\mathrm{d}x$? I've read many question/answer threads here on SE re: justification for the algebraic manipulation of $\mathrm{d}y/\mathrm{d}x$ in the standard formulation of calculus. I worked up my own shot at a justification using the definition of the derivative and would like to get the input of others whether it holds. It doesn't involve pushforward maps, non-standard analysis (two things I don't yet know), or any other tool than simple limit algebra/laws, the definition of the derivative, and (implicitly), the chain rule that makes the substitution possible. If it does hold as a justification as far as it goes, i.e., in one-dimension, it would seem to me, at least, to be a simple and clear justification in the form of an example that could be useful to other beginners. If goes wrong somewhere please let me know for my own understanding. 
Integrate $\int \frac{x}{1+x^2}\mathrm{d}x$.
We'll do this by substitution. We want to restate the original problem in terms of $u$. That means
that we want to state $x$, $1+x^2$, and $\mathrm{d}x$ in terms of $u$.
Let $u=1+x^2 \rightarrow \frac{\mathrm{d}u}{\mathrm{d}x}=2x \rightarrow \mathrm{d}x=\frac{\mathrm{d}u}{2x}$.
Let's justify that last step as that's where the justification (that I speak of) is required:
By definition, $\frac{\mathrm{d}u}{\mathrm{d}x}=2x$ is defined as
$$
\lim_{\Delta x \to 0}\left( \frac{u(x+\Delta x)-u(x)}{\Delta x} \right)=2x
$$
Recall that the limit of a quotient is equal to the quotient of the
limits as long as the denominator is not equal to zero, thus the
foregoing becomes
$$
\frac{\lim_{\Delta x \to 0}\left( u(x+\Delta x)-u(x)
\right)}{\lim_{\Delta x \to 0} \Delta x } =2x
$$
Multiplying both sides by $\lim_{\Delta x \to 0}\Delta x$ and then dividing
both sides by $2x$ yields
$$
\frac{\lim_{\Delta x \to 0}\left( u(x+\Delta x)-u(x)
\right)}{2x} =\lim_{\Delta x \to 0} \Delta x
$$
which, in the limit, is equal to
$$
\frac{\mathrm{d}u}{2x}=\mathrm{d}x
$$
Rearranging we have
$$
dx=\frac{\mathrm{d}u}{2x}
$$
as desired.
 A: The problem is to define $dx$ and $du$... so writing $dx = \frac{du}{2x}$ is non-trivial, and it's not easy to give a rigourous meaning to both of the terms. Sure, as a mnemonic, $\frac{du}{dx} = 2x \Leftrightarrow dx = \frac{du}{2x}$ is fine. but, I repeat, what is exactly $dx$?
What's easier to show is the substitution rule directly. As $\left(f(g(x))\right)' = g'(x)f'(g(x))$ we have 
$$\int_{f(a)}^{f(b)} g(x) dx = \int_{a}^{b} g(f(t))f'(t) dt$$
Indeed, let's call $G(x) = \int_0^x g(t) dt$. We have then (when g is continous) $G'(x) = g(x)$ and
$$ \int_a^b g(f(t))f'(t) dt =\int_a^b G'(f(t))f'(t) dt $$
We now use the chain rule. as $G'(f(t))f'(t) = (G(f(t)))'$, it imply
$$ \int_a^b G'(f(t))f'(t) dt = \int_a^b (G(f(t)))' dt = G(f(b))-G(f(a)) = \int_{f(a)}^{f(b)} g(x) dx $$
So the substitution rule is not hard to demonstrate. It's harder to remember, and here $dx = \frac{du}{2x}$ can be usefull as a mnemonic. If you want to give a rigourous meaning to this expression, you'll have some work to do (and definitions to create). Also remember that for a beginner, (with Riemann integration) is a notation of 
$$\int_a^b f(t) dt = \lim_{n\to +\infty} \sum_{i=0}^{n-1} f(x_i)(x_{i+1}-x_i)$$ 
When the limit is the same for all subdivisions. We could have noted it $I(a,b,f)$, and the substituting rule would be 
$$I(f(a),f(b),g) = I(a,b,f'\times(g\circ f) )$$
Here there is no $dt$
