Differentiability of a function

[Felafel]

Homework Statement

I have the function f, defined as follows:
f=0 if xy=0
f= $xysin(\frac{1}{xy})$ if $xy \neq 0$

Study the differentiability of this function.

The Attempt at a Solution

there are no problems in differentiating the function where $xy\neq0$.
the partials in (0,0) are both zero.
If i consider xy as a single variable i get from the definition of differential that:
$\lim_{xy \to 0} \frac{f(xy)-f(0)-\nabla f(0)(xy)}{\sqrt{(xy)^2-0}}$=$\lim_{xy \to 0} \frac{xy sin(1/xy)}{xy}$ doesn't exist

On the other hand, if i consider x,y separately i get:
$\lim_{(x,y) \to (0,0)} \frac{f(x,y)-f(0,0)-\nabla f(0,0)(x,y)}{\sqrt{x^2+y^2}}$=$\lim_{xy \to 0} \frac{xy sin(1/xy)}{\sqrt{x^2+y^2}}=0$ and thus the function is differentiable even when x,y are 0.

these two results are conflicting. what am i doing wrong?

 

[mfb]

The results are not conflicting. The existence of the partial derivatives is not sufficient to have a differentiable function.

 

[Felafel]

The results are not conflicting. The existence of the partial derivatives is not sufficient to have a differentiable function.

yes, i know that, but ny problem is that the differential doesn't exist if i consider xy as a single variable, but it does if i consider x and y separately, please look at the two different applications of the definition of differentiability i wrote

 

[PeroK]

I don't think that the partial derivatives exist (except trivially for x = 0 and y = 0). E.g.

$\frac{\partial{f}}{\partial{x}}|_{(0,y)} \ = \lim_{x \rightarrow 0}\frac{xysin(\frac{1}{xy})}{x} = \lim_{x \rightarrow 0}ysin(\frac{1}{xy})$

Which doesn't exist.

 

[Felafel]

should I deduce than that it is differentiable only in (0,0)?

 

[PeroK]

Yes, it's differentiable at (0, 0) but nowhere else on the x- and y-axes.

 

[mfb]

yes, i know that, but ny problem is that the differential doesn't exist if i consider xy as a single variable, but it does if i consider x and y separately, please look at the two different applications of the definition of differentiability i wrote

You did not use the definition of differentiability in (x,y) at all. If the function would be differentiable at (0,0), both derivative attempts would have to work. They do not, showing your function is not differentiable at that point.

 

[PeroK]

Why is the function not differentiable at (0,0)?

 

[mfb]

This post (and #11 and #13) is wrong, see post 15

It is not even continuous.

Consider (x,y)=(h,h) with h≠0. Then $f(h,h)=h^2 \sin(\frac{1}{h^2}) > h^2 \frac{1}{2h^2} = \frac{1}{2}$ using sin(x)>x/2 for small x>0.
Clearly the limit for $h \to 0$ cannot be 0, but f(0,0) is zero.

 

[PeroK]

I'm not sure about $sin(x) > 1/(2x)$ given that $|sin(x)| \le 1$

As (x, y) -> (0,0), $\frac{xy}{\sqrt{x^2+y^2}} → 0$

 

[mfb]

I'm not sure about $sin(x) > 1/(2x)$ given that $|sin(x)| \le 1$

I don't see the connection of your global upper bound to my local lower bound (valid at least up to pi/2).
Plot

I should have specified x>0 in "sin(x)>1/(2x) for small x", but that is not an issue as h^2 is always positive (I edited it in my previous post now).

As (x, y) -> (0,0), $\frac{xy}{\sqrt{x^2+y^2}} → 0$

Where does the denominator come from?

 

[PeroK]

Ok, so take x = 0.1:

1/2x = 1/0.2 = 5

And sin(anything) can't possibly be > 5

 

[mfb]

Sorry, my mistake. I meant sin(x)>x/2 (for 0<x<pi/2). The plot and the application to calculate the limit were correct.

 

[PeroK]

It is not even continuous.

Consider (x,y)=(h,h) with h≠0. Then $f(h,h)=h^2 \sin(\frac{1}{h^2}) > h^2 \frac{1}{2h^2} = \frac{1}{2}$ using sin(x)>x/2 for small x>0.
Clearly the limit for $h \to 0$ cannot be 0, but f(0,0) is zero.

$\lim_{h→0}h^2 \sin(\frac{1}{h^2}) = 0$

If h is getting small and |sin| is bounded by 1, then that limit must be zero, surely?

$1/h^2$ is getting very large, not very small

 

[mfb]

You are right, the function is differentiable. Sorry for the confusion.

I really need that free weekend now...

 

[Felafel]

thank you very much :)