# prove a function is continuously differentiable

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

$f$ 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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1
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. – DonAntonio Oct 28 '12 at 4:38

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I have no idea what your inequalities and bounds for $x$ and $y$ represent. Please fix those yourself. – EuYu Nov 1 '12 at 6:17
So...are you dividing by $0$ in there? That's...bad. – Cameron Buie Nov 1 '12 at 6:21
that's a function with different values in different domains – Frank Xu Nov 1 '12 at 6:53
Ah, I see. Please tell me if my interpretation is right. – EuYu Nov 1 '12 at 7:06

## 1 Answer

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 all – Frank  Xu Nov 1 '12 at 7:05
could you give me more hint? how can I apply implicit function? Is there any conclusion of implicit function about continuously differentiable? – Frank  Xu Nov 1 '12 at 7:58
– Fabian Nov 1 '12 at 10:39
never say something is obvious. when you say something is obvious, it could be not obvious – Frank  Xu Nov 1 '12 at 17:38
the part you cited as obvious is exactly the hard part – Frank  Xu Nov 1 '12 at 17:39
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