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I want to show that $f(x,y) = \sqrt{|xy|}$ is not differentiable at $0$.

So my idea is to show that $g(x,y) = |xy|$ is not differentiable, and then argue that if $f$ were differentiable, then so would $g$ which is the composition of differentiable functions $\cdot^2$ and $g$.

But I'm stuck as to how to do this. In the one variable case, to show that $q(x) = |x|$ is not differentiable, I can calculate the limit $\frac{|x + h| - |x|}{h}$ as $h\to 0^+$ and $h\to 0^-$, show that the two one-sided limits are distinct, and conclude that the limit $$\lim_{h\to 0}\frac{|x + h| - |x|}{h}$$ does not exist.

The reason this is easier is that I do not have to have in mind the derivative of the function $q$ in order to calculate it.

But in the case of $g(x,y) = |xy|$, to show that $g$ is not differentiable at $0$, I would need to show that there does not exist a linear transformation $\lambda:\mathbb{R}^{2}\to\mathbb{R}$ such that

$$\lim_{(h,k)\to (0,0)}\frac{\left||hk| - \lambda(h,k)\right|}{|(h,k)|} = 0$$

I thought of assuming that I had such a $\lambda$, and letting $(h,k)\to (0,0)$ along both $(\sqrt{t},\sqrt{t})$ and $(-\sqrt{t},-\sqrt{t})$ as $t\to 0^{+}$, but this didn't seem to go anywhere constructive.

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Seems to me that your insight about $|xy|$ is not correct. The one-variable analog of this function seems to me to be $x^2$, not $|x|$. Think of what the graph $z=xy$ looks like: it’s very similar to $z=x^2-y^2$, but rotated $45^\circ$ about the $z$-axis. As a result, I am pretty sure that $|xy|$ is differentiable at the origin, even though it’s not at all other points on the $x$ and $y$ axes. –  Lubin Aug 14 '12 at 0:30
    
Since $|xy| \leq \|(x,y)\|^2$, it follows that $(x,y) \to |xy|$ is differentiable at the origin (with derivative $0$). –  copper.hat Aug 14 '12 at 0:54

3 Answers 3

up vote 4 down vote accepted

Note: My previous answer was incorrect, thanks to @Lubin for catching that.

Simplify your life and show that $\phi(x) = f(x,x) = |x|$ is not differentiable at $0$. It will follow from this that $f$ is not differentiable at $0$.

Look at the definition of differentiability for this case, which is that $\lim_{h \to 0, h \neq 0} \frac{\phi(h)-\phi(0)}{h} $ exists. We have $\phi(0) = 0$, and $\phi(h) = |h|$, so we are looking at the limit of $h \mapsto \mathbb{sign}(h)$ as $h \to 0$.

If you choose $h_n = \frac{(-1)^n}{n}$, it is easy to see that $\frac{\phi(h_n)-\phi(0)}{h_n} = (-1)^n$, hence it has no limit. It follows that $f$ is not differentiable at $0$.

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I don’t agree that the nondifferentiability of $f$ at $(0,1)$ has anything to do with the fact of nondifferentiability of $f$ at $(0,0)$. –  Lubin Aug 14 '12 at 0:34
    
@Lubin: You are correct, I must have been asleep... –  copper.hat Aug 14 '12 at 0:42
    
@Lubin: I have fixed my proof bug. –  copper.hat Aug 14 '12 at 0:52
    
you weren’t nearly as asleep as I was when I was in the midst of a many-hundred-word response. We are in total agreement now. –  Lubin Aug 14 '12 at 1:07
    
It's worse that that. I typed up the above first, and before saving the edits I thought could make non-differentiability even more obvious by taking $y=1$ (to get $\sqrt{|x|}$) and rewrote my answer. Oh well, I shouldn't pack bags and answer questions at the same time :-). –  copper.hat Aug 14 '12 at 1:11

Use directional derivatives. Note that the limit of $$\frac{f(h,h)-f(0,0)}{h\sqrt{2}}=\frac{|h|}{h\sqrt{2}}$$ as $h\to 0^+$ is different than that as $h\to 0^-$.

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In agreement with @Cameron, I would show the nondifferentiability of $f(x,y)$ at $(0,0)$ in the following way. Intersect the graph with a vertical plane through the $z$-axis, say given by $y=\lambda x$. The intersection is given by $z=|\lambda|^{1/2}\cdot|x|$. So above each of the four quadrants of the $x,y$-plane, the graph consists of strings stretched out from the origin at varying angles. In particular, the “diagonal” plane $y=x$ intersects the graph in a V-shaped figure exactly like the familiar graph of absolute value in one-variable calculus.

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