# Taking partial derivative in polar coordinates

1. May 11, 2014

### nigels

In a problem that requires converting from cartesian to polar coordinates, I need to take $\frac{dr}{dx}$. I tried doing it two different ways but getting two completely different answers..

Method 1:

$r=\sqrt{x^2+y^2}$

$\frac{dr}{dx}=\frac{1}{2}\frac{1}{\sqrt{x^2+y^2}}2x \;\; = \frac{1}{r}r\cos\theta \;\; = \;\; \cos\theta$

Method 2:

$x=r\cos\theta$

$r=\frac{x}{\cos\theta}$

$\frac{dr}{dx} = \frac{1}{\cos\theta}$

I know that Method 1 gives the correct answer. But, having not done derivations in a while, what's wrong with my steps in Method 2? Thanks a lot for your help!!!

2. May 11, 2014

### NihilTico

I claim that theta is a function of x and y, can you find what theta is then?

Last edited: May 11, 2014