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?

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    $\begingroup$ Related: math.stackexchange.com/questions/21199/is-fracdydx-not-a-ratio $\endgroup$ – user122283 Jun 6 '14 at 1:52
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    $\begingroup$ Actually, nonzero derivative does mean that $x$ is a function of $y$ when both are restricted to a little rectangle, and indeed $x$ is also differentiable. In more variables this is called the Inverse Function Theorem, and a related version the Implicit Function theorem. $\endgroup$ – Will Jagy Jun 6 '14 at 2:00
  • $\begingroup$ @WillJagy please see math.stackexchange.com/questions/814308/… for the example where derivative is nonzero but function is still not invertible $\endgroup$ – David Jun 6 '14 at 2:34
  • $\begingroup$ @David, anything can go wrong if you weaken hypotheses too far. From reading this question, i did not think that you were saying your function was differentiable at only a single point...there are better books for exploring pathologies. I would say that Kaplan is mostly about what to do when things are working well. see store.doverpublications.com/0486428753.html $\endgroup$ – Will Jagy Jun 6 '14 at 2:38

I think a problem here is that you are blending together the original function $y=y(x)$ and its local linearized version $dy=2dx$. This last just means that near the point $(1,2)$, the function $y=y(x)$ changes like the function $y=2x$ changes. y=y(x) is the original function and $dy=2dx$ is its linearization/differential. They are two different functions; one is not necessarily linear and a function of x, but the differential is, by construction, linear (in dx), and a function of $dx$.


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