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The function is as follows: $$f(x,y)=\sqrt{\left |xy \right |} $$ I have to check whether it is continuous, differentiable and has defined partial derivatives at $(0,0)$. My attempt is as follows:

  1. Function is discontinuous at the origin.

  2. Not differentiable at the origin because of the pointy peak (haha); and

  3. Partial derivatives are as follows: $f_x=\frac{\sqrt y}{2\sqrt x}$ and $f_y=\frac{\sqrt x}{2\sqrt y}$ and are not defined at origin.

Is my reasoning correct?

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    $\begingroup$ Why is the function discontinuous at the origin? $\endgroup$
    – gangrene
    Nov 21, 2013 at 15:19
  • $\begingroup$ Because modulus functions are discontinuous at the origin...? :) $\endgroup$
    – Artemisia
    Nov 21, 2013 at 15:24
  • $\begingroup$ That's not true. Maybe you are confusing yourself with the fact that the function $f:\mathbb{R} \to \mathbb{R}$, $f(x)=|x|$ is not differentiable at the origin... $\endgroup$
    – gangrene
    Nov 21, 2013 at 15:28
  • $\begingroup$ Ah maybe... so the function is continuous? Hmmm. And differentiable? $\endgroup$
    – Artemisia
    Nov 21, 2013 at 15:30
  • $\begingroup$ Be careful with partial derivatives, your formulae are wrong! $\endgroup$
    – Siminore
    Nov 21, 2013 at 16:03

2 Answers 2

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To show that the function is continuous at the origin, you may use the polar coordinates: putting $x=\rho \cos \theta$ and $ y=\rho \sin \theta$ with $\rho >0$ and $\theta \in [0, 2 \pi)$, you have that $$\sqrt{|xy|}=\sqrt{\rho^2 |\cos \theta \sin \theta|}\le \rho \to 0$$if $\rho \to 0$. About the differentiability, I think that you are right (anyway it is better if you do a background check with the definition).

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Since $\sqrt{|x|} \to 0$ as $x \to 0$, it follows immediately that $$ \sqrt{|xy|} = \sqrt{|x|} \sqrt{|y|} \to 0 $$ as $|x|+|y| \to 0$. Hence $f$ is continuous at the origin. Since $f=0$ on both coordinate axes, the two partial derivatives do exist at the origin, and are both zero. Can you study the issue of differentiability by yourself? Hint: check that the directional derivative along the direction $\mathbf{v}=(1,1)$ does not exist at the origin.

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  • $\begingroup$ So the function is not differentiable. $\endgroup$
    – Artemisia
    Nov 22, 2013 at 14:33
  • $\begingroup$ No, it isn't differentiable. $\endgroup$
    – Siminore
    Nov 22, 2013 at 15:36

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