Is \( f(x,y) = \sqrt{|xy|} \) Differentiable at \( (0,0) \)?

[Buri]

Here's the problem:

Let f(x,y) = sqrt(|xy|) if x >= 0 OR -sqrt(|xy|) if x < 0. Show that all directional derivatives in the direction of h exist, but f is not differentiable at (0,0).

Here's what I've done:

Basically all I need to show is the directional derivative is not linear. Since if the differential exists then df_(0,0)(v) = D_v f(0,0), but the directional derivative will turn out to be not linear and hence it would contradict a result I have.

So I have that for h = (h1,h2) that the following exists:

lim [t->0] (1/t)[f((0,0) + t(h1,h2)) - f(0,0)]
= lim [t->0] (1/t)sqrt(|t²h1h2|)

But here's where I end up getting:

lim [t->0] [|t|sqrt(|h1h2|)]/t

The above is for th1 > 0. But this limit doesn't exist as when I take it from the left and then right I end up with the same expression but with a negative in front of one of them :S Can anyone help me out on this?

Thanks!

 

[HallsofIvy]

Suppose we take the derivative in a direction with angle $\theta$ to the x-axis and let $m= tan(\theta)$. Then your (h1, h2) can be written (h, mh). Now we have $f(h, mh)- f(0, 0)= \sqrt{mh^2}= \sqrt{m}|h|$ for h positive and $-\sqrt{m}|h|$ for h negative. Of course, if h> 0 then |h|/h= 1 and if h< 0, |h|/h= -1 so we have $(f(h, mh)- f(0,0))/h= \sqrt{m}$ in either case. Since that is a constant, its limit, as h goes to 0, is just that same constant, $\sqrt{m}$. That shows that the directional derivative exists.

A function, f(x, y) is "differentiable" at (0, 0) if and only if
$$f(x,y)= f(0, 0)+ \frac{\partial f}{\partial x}(0, 0)x+ \frac{\partial f}{\partial y}(0, 0)y+ \epsilon(x,y)$$
where
$$\lim_{(x,y)\to 0}\frac{\epsilon(x,y)}{\sqrt{x^2+ y^2}}= 0$$.

Here, it is easy to see that the partial derivatives as both 0 at (0, 0) so we must have $\epsilon(x,y)= f(x,y)$.

 

[Buri]

Suppose we take the derivative in a direction with angle $\theta$ to the x-axis and let $m= tan(\theta)$. Then your (h1, h2) can be written (h, mh). Now we have $f(h, mh)- f(0, 0)= \sqrt{mh^2}= \sqrt{m}|h|$ for h positive and $-\sqrt{m}|h|$ for h negative. Of course, if h> 0 then |h|/h= 1 and if h< 0, |h|/h= -1 so we have $(f(h, mh)- f(0,0))/h= \sqrt{m}$ in either case. Since that is a constant, its limit, as h goes to 0, is just that same constant, $\sqrt{m}$. That shows that the directional derivative exists.

I'm confused. For h positive you wrote:

$f(h, mh)- f(0, 0)= \sqrt{mh^2}= \sqrt{m}|h|$

But then why is it that you don't take the limit as h goes from the right and then from the left of this one? To see if the right/left handed side limits exist?

Wait I guess the one h > 0 and the other for h < 0 are basically the right and left handed sided limits already? I guess I shouldn't have to "divide" them again.

And another question. I honestly don't understand anything of the rest since I have no idea what partial derivatives are yet. We've only covered the direction derivative and the the 'normal' derivative. But is my idea on how to finish the question okay? D_h f((0,0)) is non-linear, so the differential at h won't be able to equal D_h f(0,0) since the differential is supposed to be linear.

Thanks for your help.