Partial Derivatives of Cartesian and Polar Coordinate Systems Plane polar coordinates, $r$ and $\phi$, and Cartesian coordinates, $x$ and $y$, are related by 
$$x=r\cos\phi,\ y=r\sin\phi,$$
and
$$r^2=x^2+y^2,\ \phi=\tan^{-1}({y \over x}).$$

To try to find the partial derivative $\partial r \over \partial x$, I used two approaches:


Using $x=r \cos \phi$, 
$${\partial x \over \partial r}=\cos \phi,$$
and using ${\partial x \over \partial r}=({\partial r \over \partial x})^{-1},$
$${\partial r \over \partial x}=\frac{1}{\cos \phi}.$$
Using $r^2=x^2+y^2$,
$$r=(x^2+y^2)^{1/2},$$
$$\frac{\partial r}{\partial x}=\frac{x}{(x^2+y^2)^{1/2}}=\cos\phi.$$


Why are the two answers different? I think the second approach gives the correct answer but I do not understand what is wrong with the first approach.
Division in partial derivatives is just a notation so $\frac{\partial r}{\partial x} \neq \frac{\partial x}{\partial r}^{-1}$.
What you are doing using the "inverse" of the derivate, is a property that has the derivation when proceding with bijective functions, this means that they have and inverse function so the derivative of the inverse function is the inverse of the derivative of the initial function. Let me suggest you to read these pages:


Bjective function: https://en.wikipedia.org/wiki/Bijection
Inverse function theorem: https://en.wikipedia.org/wiki/Inverse_function_theorem


This is, the function $r^2 = x^2 + y^2$ is not bijective, so you can not do the derivative as you proposed in the second case.