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So, I have $ f(x,y) = (x^2-y^2, 2xy) $, which is a local $\mathcal C^1$ isomorphism in $\mathbb R^2 \setminus \{(0,0)\}$.

I have to write this function in polar coordinates:

$$f(x,y) = f(r\cos\phi, r\sin\phi).$$

My beginnings: I know that

$$df(r, \phi) = \cos\phi -r\sin\phi \sin\phi r\cos\phi.$$

But I really have no clue how to work this one out. It is a single exercise of this kind in my book; so, I probably don't need it for the test, but I would like to know.

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    Just plug in $r\cos \phi$ and $r\sin\phi$ for $x$ and $y$ respectively. Note that trig identities will allow you to simplify your result.2012-01-06
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    As @Aaron suggested $f(r,\theta)$ can be obtained by replacing $x$ with $r\cos{\theta}$ and $y$ with $r\sin{\theta}$2012-01-06

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