Taking partial derivative in polar coordinates

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

Method 1:

[itex]r=\sqrt{x^2+y^2}[/itex]

[itex]\frac{dr}{dx}=\frac{1}{2}\frac{1}{\sqrt{x^2+y^2}}2x \;\; = \frac{1}{r}r\cos\theta \;\; = \;\; \cos\theta[/itex]


Method 2:

[itex]x=r\cos\theta[/itex]

[itex]r=\frac{x}{\cos\theta}[/itex]

[itex]\frac{dr}{dx} = \frac{1}{\cos\theta}[/itex]


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