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?

• – user122283 Jun 6 '14 at 1:52
• 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. – Will Jagy Jun 6 '14 at 2:00
• @WillJagy please see math.stackexchange.com/questions/814308/… for the example where derivative is nonzero but function is still not invertible – David Jun 6 '14 at 2:34
• @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 – 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 $dx$.