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Hello everyone I have this problem, Can somebody help me with this?

$f:\mathbb{R}^2\rightarrow{}\mathbb{R}$ is defined by:

$$f(x,y) = \left \{ \begin{matrix} \ln\left(\displaystyle\frac{x}{y}\right) & \mbox{if } xy\geq{0} \\ 0 & \mbox{if }xy<0\end{matrix}\right. $$

Is $f$ a differentiable function?

Thanks for your help :D , have a nice day

Shuhao Cao
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user63192
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    What is the value of the function at the point $(0,0)$? How about $(1,0)$? And $(0,1)$? – Calvin Lin Jun 03 '13 at 04:44
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    The function behaves well (according to your definition) except on the axes. On these points, the function is clearly differentiable. However, at the axes, your function definition doesn't make sense, and so we can't talk about it being differentiable. – Calvin Lin Jun 03 '13 at 04:48
  • @CalvinLin :o Thanks for your time and your answer, it help me a lot :D – user63192 Jun 03 '13 at 04:54
  • @CalvinLin For example if $x=0,\ y\neq 0$ we have ln(0) (doesn´t exist), if $x\neq 0,\ y=0$ we have ln(x/0) (doesn´t exist), at the point (0,0) the same case doesn´t exist – user63192 Jun 03 '13 at 05:16

2 Answers2

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The function behaves well (according to your definition) except on the axes. On these points, the function is clearly differentiable. However, at the axes, your function definition doesn't make sense (the values do not exist), and so we can't talk about it being differentiable.

Calvin Lin
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  • $f$ isn´t a continuous function then $f$ can´t be differentiable.

    Thanks you very much

     – user63192 Jun 03 '13 at 04:59
  • You're missing the point. It is not defined at those points at all. Consider the function $g(x) = \frac{1}{x}$. It is not defined at 0, so we can't say anything about the function at 0. – Calvin Lin Jun 03 '13 at 05:01
  • For example if $x=0,\ y\neq 0$ we have ln(0) (doesn´t exist), if $x\neq 0,\ y=0$ we have ln(x/0) (doesn´t exist), at the point (0,0) the same case doesn´t exist – user63192 Jun 03 '13 at 05:15
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Besides to @Calvin sufficient and strong deduction,the function as you defined it, is not continuous at least at the origin $(0,0)$. Take the path $y=mx$ for various $m$. This can be extended to other problematic points.

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Mikasa
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