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

$f$ 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.


  • 1
    $\begingroup$ 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. $\endgroup$ – DonAntonio Oct 28 '12 at 4:38
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    $\begingroup$ I have no idea what your inequalities and bounds for $x$ and $y$ represent. Please fix those yourself. $\endgroup$ – EuYu Nov 1 '12 at 6:17
  • $\begingroup$ So...are you dividing by $0$ in there? That's...bad. $\endgroup$ – Cameron Buie Nov 1 '12 at 6:21
  • $\begingroup$ that's a function with different values in different domains $\endgroup$ – Frank Xu Nov 1 '12 at 6:53
  • $\begingroup$ Ah, I see. Please tell me if my interpretation is right. $\endgroup$ – EuYu Nov 1 '12 at 7:06

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


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 $\partial_y f$?


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.

  • $\begingroup$ doesn't help at all $\endgroup$ – Frank Xu Nov 1 '12 at 7:05
  • $\begingroup$ could you give me more hint? how can I apply implicit function? Is there any conclusion of implicit function about continuously differentiable? $\endgroup$ – Frank Xu Nov 1 '12 at 7:58
  • $\begingroup$ Did you take a look at the theorem on wikipedia. 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. $\endgroup$ – Fabian Nov 1 '12 at 10:39
  • $\begingroup$ never say something is obvious. when you say something is obvious, it could be not obvious $\endgroup$ – Frank Xu Nov 1 '12 at 17:38
  • $\begingroup$ the part you cited as obvious is exactly the hard part $\endgroup$ – Frank Xu Nov 1 '12 at 17:39

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