I am reading the book "Advanced Calculus" written by Kaplan, and here is what I have:
Suppose that $y(x)$ is a differentiable function at $x = x_0$. Then, we can write $y(x_0+\Delta x) = y(x_0) + y'(x_0)\cdot \Delta x + \epsilon\cdot \Delta x, \, where \, \epsilon \rightarrow 0 \: as \, \Delta x \rightarrow 0$. We denote the differential of $y$ by $dy = y'(x_0)\cdot\Delta x$.
Now note that if $g(x) = x$, then $g$ is clearly differentiable at $x = x_0$, so we have $dg = dx = 1\cdot\Delta x$, from which it follows that $dy = y'(x_0)\cdot\Delta x= y'(x_0)\cdot dx$
Also, if $y'(x_0) \neq 0 $, we can write $dx = \dfrac{1}{y'(x_0)}dy$
I am uncomfortable with the last line $dx = \dfrac{1}{y'(x_0)}dy$ because I think that in order for this expression to be "meaningful", it must be true that x can be written as a function of y, and this function must be differentiable at $y = f(x_0)$.
But, as far as I know, mere existence of nonzero derivative of $f(x)$ at $x=x_0$ does not imply the invertibility of y(x), let alone its differentiability.
To illustrate my point, say, for instance, that we have $y(2) = 1$ and $y'(2)= 3$ so that $dy = 3dx$ Then, according to Kaplan, we can write $dx = \dfrac{1}{3}dy$. But then, if somone else sees only $dx = \dfrac{1}{3}dy$ at $(x,y) = (1,2)$, then he/she would think that x is a function of y with $x(1) = 2$ and $x'(1) = \dfrac{1}{3}$. But, the fact that $y(2) = 1$ and $y'(2) = 3$ does not imply $x(1) = 2$ and $x'(1) = \dfrac{1}{3}$. Right?
How can you freely move around $dy$ and $dx$ as if it does not matter which variable is an independent varible?
