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I need help with a question that appeared in my test.

True or False:

Let $f$ be a function $f:\mathbb{R^2} \to \mathbb{R}$ not differentiable at (0,0), then $f^2$ is not differentiable at (0,0).

I answered False. I gave an example of $\sqrt{x+y-1}$ which isn't defined at (0,0) let alone differentiable, yet $x+y-1$ does. But my professor wrote on my test that $\sqrt{x+y-1}$ is differentiable at (0,0).

who among us is incorrect?

The Chaz 2.0
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Illya Broder
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    Your function is not a function $\mathbb{R}^2 \to \mathbb{R}$, but you can construct one with values $1$ and $-1$ only. – njguliyev Oct 17 '13 at 16:51
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    Your professor might not be literally sticking to the definitions (I see no point in discussing differentiability of a function outside the interior of its domain), but your answer doesn't qualify as a counterexample; you've sidestepped the scope, intent (and crux) of the question. – Jonathan Y. Oct 17 '13 at 17:03
  • What is your professor smoking? Your example is undefined at $(0,0)$ so it can't be be differentiable there. The real reason your example is bad is because your $f(0,0)$ is undefined so $f(0,0)^2$ is undefined. – Stefan Smith Oct 18 '13 at 02:39

2 Answers2

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$f(x,y)=|x|$ may be a counter example.

Myshkin
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The function you specified is not defined on $\mathbb{R}^2$.

Try $f(x,y) = \sqrt{|xy|}$. Then $f$ is not differentiable at $(0,0)$, but $f(x,y)^2 = |xy|$ is differentiable at $(0,0)$ (since $|xy| \le \frac{1}{2} (x^2+y^2)$).

copper.hat
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