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$f(x,y) =\begin{cases}\arctan(y/x) & x\neq 0\\ \pi/2 & x=0,y>0\\-\pi/2 & x=0,y<0.\end{cases}$

$f$ is defined on $\Bbb R^2\smallsetminus\{(0,0)\}.$

Show that $f$ is continuously differentiable on all of its domain.

Also use implicit function to show the above proof again.

Thanks!

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    Ah, I see. Please tell me if my interpretation is right.2012-11-01

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Maybe rewriting your equation as $ x \tan f = y$ does help?

Edit:

Given the fact that the first hint did not help. Here, is the second hint: you can rewrite your equation as $ F(x,y,f) = x \tan f - y =0.$ Can you then use the implicit function theorem to learn something about $\partial_x f$ and $\partial_y f$?

Edit2:

I just did see that you have changed your question and thus the points with $f=\pi/2 + n\pi$ are not excluded any more. In this case you should rewrite your equation as (check the special points at $f=\pi/2 + n\pi$ separately) $F(x,y,f) = x \sin f - y \cos f =0.$ and then apply the implicit function theorem.

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    @FrankXu: I thought discussed this already. I assume you know that the product/sum of two continuously differentiable functions is continuously differentiable. You agreed to know that $\sin f$ and $\cos f$ are continuously differentiable. I did not ask you if you know that $x$ and $y$ are continuously differentiable (because I thought you know it). Is you question why $x$ (or for that matter $y$) is a continuously differentiable function? Because as soon as this is clear you should agree that $F(x,y,f) = x \sin f - y \cos f$ is continuously differentiable in all its variables.2012-11-03