Contradiction for $(dy/dx)^{-1}=dx/dy$

Question

So I have seen in the derivation of Euler chain Equation (i.e. $\displaystyle \frac{\partial x}{\partial y} \cdotp \frac{\partial y}{\partial z} \cdotp \frac{\partial z}{\partial x} \ =\ -1$ )

where we use $\displaystyle \frac{\partial y}{\partial x} \ =\ \frac{1}{\partial x/\partial y}$

So, I have a contradiction.

Let's say $x = rcos(\theta)$ -------- (1)

and $y = rsin(\theta)$, so we have $x^2+y^2=r^2$ ---------(2)

$\displaystyle \frac{\partial r}{\partial x} \ = cos(\theta)$ from eqn(2)

And $\displaystyle \frac{\partial x}{\partial r} \ = cos(\theta)$ from eqn(1).

Where is the problem?

If $\displaystyle \frac{\partial y}{\partial x} \ =\ \frac{1}{\partial x/\partial y}$ is false, then what are the conditions where it's true?

If $x=r\cos \theta$ then $r=\frac x{\cos \theta}$ so $\frac {\partial r}{\partial x}=\frac 1{\cos \theta}$. — lulu, Sep 23 '23 at 12:13
Not sure if a duplicate, but this is closely related to Sufficient conditions for partial derivative reciprocal. — Mark S., Sep 23 '23 at 13:34

score 2 · Accepted Answer · edited Sep 23 '23 at 13:41

Here, you have functions of two variables, and when there is ambiguity, you have to indicate which variable is fixed when you compute a partial derivative.

For example, you can fix $\theta$ when you derive $x=r\cos\theta$ and get $\displaystyle \Big( \frac{\partial r}{\partial x} \Big)_\theta\ = \frac{1}{cos(\theta)}.$

Or you can fix $y$ when you derive $x^2+y^2=r^2$ and get $\displaystyle 2x \Big(\frac{\partial x}{\partial r} \Big)_y = 2r$, so $\displaystyle \Big(\frac{\partial x}{\partial r} \Big)_y = \frac{r}{x} = \frac{1}{cos(\theta)}.$

Contradiction for $(dy/dx)^{-1}=dx/dy$

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