Proposition 1 is an example of the situation when $\mathrm dy\over\mathrm dx$ can't be seen as a quotient.
(the definition of differential and derivative is at the end)
Proposition 1:
Forall function $f$ satisfied with $x=f(t)$, if its derivative at $t_0$ is equal to $\mathrm dx\over\mathrm dt$, $g$ satisfied with $y=g(x)$ is also derivable at $f(t_0)$, then the derivative of $g\circ f$ at $t_0$ is $${\mathrm dy\over\mathrm dt}={\mathrm dy\over \mathrm dx}\cdot{\mathrm dx\over\mathrm dt}$$
(End of proposition.)
The conclusion is that least one of ${\mathrm dy\over \mathrm dx}$ and ${\mathrm dx\over \mathrm dt}$ can not be understood as quotient. Below is the explanation. If both of them are quotient, then $\mathrm dx$ will be cancelled out; however, in this case, chain rule is not meaningful, it is not even a mathematics statement (only a sentence without maths meaning); nevertheless, chain rule is not a meaningless sentence. Let's look at the reason for if seeing both as a quotient, then chain rule is a meaningless sentence: not all function $f$ which has a derivative at $x_0$ is satisfied with the statement, for $\mathrm dx$ can be $0$ which can not be placed at denominator; furthermore, the quantity with the same notion (such as $\mathrm dx$ in that sentence) is seen as identical in maths. Therefor, there is at least one of them can not be seen as quotient.
Proposition 2 is an example of $\mathrm dy\over\mathrm dx$ can only be seen as a quotient.
Proposition 2:
Suppose function $f$ satisfied with $x=f(t)$ is derivable at $t_0$, $g$ satisfied with $y=g(u)$ is also derivable at $u_0$, then$${\mathrm dx\over\mathrm dt}\cdot{\mathrm dy\over\mathrm du}={\mathrm dy\over \mathrm dx}\cdot{\mathrm dx\over\mathrm du}$$
(End of proposition)
In this proposition, both of them can only be understood as quotients, otherwise, it is a fake statement or meaningless sentence. There is two situation if one understand it the other way round:
- If $y$ is not with respects to $t$, both ${\mathrm dx\over\mathrm du}$ and ${\mathrm dy\over\mathrm dt}$ are not defined, the sentence is meaningful.
- If $y$ is with respects to $t$, both ${\mathrm dx\over\mathrm du}={\mathrm dy\over\mathrm dt}=0$, the sentence is meaningful, but it is a fake statement.
It is concludeble that, both of ${\mathrm dx\over\mathrm du}$ and ${\mathrm dy\over\mathrm dt}$ in this proposition must be treated as quotient.
Definition of derivative:
Suppose $f$ of is a function defined on a subset of real number set, $D$; also, $x_0\in D$
$$\forall A(\forall\varepsilon\in\mathbf R_+\exists\delta\in\mathbf R_+\forall x\in\mathring U(x_0,\delta)(|{f(x)-f(x_0)\over x-x_0}-A|<\varepsilon)$$
$$\leftrightarrow A\mathrm{\ equals\ the\ derivative\ of}\ f\mathrm{\ at}\ x_0)$$
$A$ is denoted as $\displaystyle{\mathrm df\over\mathrm dx}$
Definition of differential:
Suppose the derivative of function $f$ at the point $x_0$ is $A$,
$$\forall g(\forall h\in\mathbf R(g(x_0,h)=Ah)\leftrightarrow g\mathrm{\ equal\ the\ differential\ of\ }f\ \mathrm{at}\ x_0)$$
$Ah$ is denoted as $\mathrm df$, called the dependent variable of differential