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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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    You're repeating *exactly* your question from 3 hours ago. You must be patient and wait until somebody deals with that, and not send over and over the same question.2012-10-28
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    I have no idea what your inequalities and bounds for $x$ and $y$ represent. Please fix those yourself.2012-11-01
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    So...are you dividing by $0$ in there? That's...bad.2012-11-01
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    that's a function with different values in different domains2012-11-01
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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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    doesn't help at all2012-11-01
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    could you give me more hint? how can I apply implicit function? Is there any conclusion of implicit function about continuously differentiable?2012-11-01
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    Did you take a look at the theorem on [wikipedia](http://en.wikipedia.org/wiki/Implicit_function_theorem#Statement_of_the_theorem). If $F$ is continuously differentiable (which should be obvious) and the Jacobian $\partial F/\partial f$ is invertible (which is just a number for us) then $f(x,y)$ is continuously differentiable.2012-11-01
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    never say something is obvious. when you say something is obvious, it could be not obvious2012-11-01
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    the part you cited as obvious is exactly the hard part2012-11-01
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    @Frank: I'm a bit confused: I believe you know that $\sin f$ and $\cos f$ are in $C^\infty$? I truly don't understand to which subtlety you are hinting to...2012-11-01
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    but you have x and y here. you have to prove it's ctsly diff when x= 02012-11-02
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    @FrankXu: $F(x=0,y,f)= - y \cos f$ I'm still not sure I understand your remark.2012-11-02
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    didn't you see x = 0 is a breaking points?2012-11-02
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    @FrankXu: no I do not see how the point $x=0$ is any special point for the function $F(x,y,f) = x \sin f - y \cos f $. I'm not sure if this discussion makes any sense so far you were not given any real indication what you are worrying about.2012-11-02
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    Because at x = 0, you have to prove F is continuously differentiable at that point.2012-11-02
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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