"Cancelling Out" Differential Forms in Chain Rule Let's think about$$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}.$$ Here function $u(x)$ is differentiable at $x=x_0$ and $y(u)$ is differentiable at $u=u_0=u(x_0)$.
I know this is a common question, but trying to learn about differential geometry and about differential forms in particular, I thought about the fundamental reason why we can't cancel out the $du$'s. I want you to check if this is correct.
Say that the tangent vector at $x=x_0$ is $\vec{a} = (a_1, a_2)$, and the tangent vector at $u=u_0$ is $\vec{b} = (b_1, b_2)$. Then the $du$ in the former is $du(\vec{a}) = a_2$, a function of $\vec{a}$, and the $du$ in the latter is $du(\vec{b}) = b_1$, a function of $\vec{b}$.
Unless it is a very special case that $a_2 = b_1$, the $du$'s have different values so we can't "cancel out" them (and I personally consider setting $a_2 = b_1$ a bit meaningless and just harms generality).
Is this logic correct?
 A: I find the entire premise of the question is a little flawed. You want to regard derivatives as ratios of the corresponding differentials, but this is simply not true (or at best a bit circular). It's important to keep in mind that $df/dx$ is just notation, and shouldn't really be regarded as a true ratio of well-defined quantities. Differential forms are concepts based on derivatives, not the other way around.
I would go as far as to say that trying to regard the chain rule as "cancellation" is not just wrong, but entirely senseless. However, it is very suggestive nonsense, so it's perhaps worth exploring why it's doomed to failure.
A slight aside first. The way you regard differential forms is not quite correct. Your functions are one-dimensional (judging from context, since you never specified domain), but you take your differential forms to be functions of two-dimensional vectors. This is not how you define differential forms. In your notation, the correct thing would be $du(a_1) = a_2$.
Just to avoid confusion, let us first formalize the concept of a differential form. In one dimension, the concept of a differential form is little more than notation. We define $df_x:\mathbb{R}\rightarrow \mathbb{R}$ as the linear map
$$df_x(r) = f'(x)r,$$
for all $r \in \mathbb{R}$. Moreover, this is an $x$-dependent linear map. Due to the fact that a linear map in one-dimension is given by a single real number (a $1\times 1$ matrix, if you will), the concept of a differential form here is completely equivalent to the derivative. You gain nothing in one dimension by working with differential forms, which is why you almost never see it used in single variable calculus. The concept of a differential form is non-trivial in higher dimensions however, because the resulting linear maps are no longer trivial.
Now, we could define ratios of differential forms to see what we get. The first thing that goes immediately wrong is that this ratio will be undefined when the form in the denominator is at a critical point. This is already damning in my opinion, but let's continue regardless. Maybe it will work everywhere the ratio is well-defined. Let's use your set up where we have functions $f$ and $u$. Let me call $g = f\circ u$. This brings us to our second issue, which is the fact that the "$df$" on the left-hand side should really be "$dg$". The chain rule written with unambiguous notation should really be
$$\frac{dg}{dx} = \frac{df}{du}\frac{du}{dx},$$
but of course this is not nearly as memorable. The mnemonic for the chain rule relies on the crucial overloading of notation, where we regard $df/dx$ not as the derivative of $f(x)$ but of $(f\circ u)(x)$. When you try to write everything down carefully and unambiguously, you find immediate issues with the ratio interpretation of derivatives. The notation $df/dx$ is not only somewhat misleading, but ultimately ambiguous. This issue is only compounded in higher dimensions with the introduction of various partial derivatives. There, you have more explicit breakdowns, like the triple product rule.
A: If
$$lim_{dx \to 0,dy\to 0}\frac{dy}{dx}= \dfrac{lim_{dy \to 0,dt\to 0}\frac{dy}{dt}} {lim_{dx \to 0,dt\to 0}\frac{dx}{dt}}$$
can be abbreviated to symbolically denote a ratio of two limits
$$\frac{dy}{dx}= \frac{dy}{dt} / \frac{dx}{dt} \tag1$$
then $$ {lim_{dy \to 0,dt\to 0}\frac{dy}{dt}} \cdot {lim_{dx \to 0,dt\to 0}\frac{dt}{dx}}$$
can as well  be abbreviated to symbolically denote a product of two limits
$$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} \tag2 $$
(1) and (2) are obtained by cancellation of differentials . They must be together correct or together wrong.
