Question regarding differentiation I have a question on differentiation.
Suppose we have $y=f(x)$.
Is ${dy\over dx}={1 \over {dx \over dy}}$?
I know we can find some counterexamples to it, like in polar coordinates:
$$x=r\cos(\theta),\ y=r\sin(\theta)$$
$${\partial x\over \partial r}=\cos(\theta)={\partial r\over \partial x}$$
This part comes from the Laplace equation in PDE:
Given the following equation:
$\Delta u=0,\ u=u(x,y)$
$Change\ x=r\cos(\theta),\ y=r\sin(\theta)$
The Jacobian of the transformation is
$$\left( \begin{array}{cc}
{\partial x \over \partial r} & {\partial y \over \partial r}\\
{\partial x \over \partial \theta} & {\partial y \over \partial \theta}\end{array} \right)=\left( \begin{array}{cc}
{\cos(\theta)} & {\sin(\theta)}\\
{-r\sin(\theta)} & {r\cos(\theta)}\end{array} \right)$$
The inverse matrix is
$$\left( \begin{array}{cc}
{\partial r \over \partial x} & {\partial \theta \over \partial x}\\
{\partial r \over \partial y} & {\partial \theta \over \partial y}\end{array} \right)=\left( \begin{array}{cc}
{\cos(\theta)} & {-\sin(\theta)\over r}\\
{\sin(\theta)} & {cos(\theta)\over r}\end{array} \right)$$
This portion comes from the book "Partial Differential Equations, an Introduction" by Walter Strauss. He also gives a warning that ${\partial r \over \partial x}\not=({\partial x \over \partial r})^{-1}$. 
I will like to know, under what circumstances can ${dy\over dx}={1 \over {dx \over dy}}$?
Thank you very much for your help.

It is true if you are dealing with a function of one variable $y = f(x)$.
