Question

Show that fx(0, 0) and fy(0, 0) both exist, but that f is not differentiable at (0, 0) where f define as f(x, y) = { (-3xy)/(x^2 + y^2) if (x, y) ≠ 0
0 if (x, y) = (0, 0)

Answer 1

By fx and fy, do you mean the partial derivatives with respect to x and y?

Answer 2

yes

Answer 3

The partial derivatives are:\[\frac{ \partial f}{ \partial x }(x, y)=-\frac{ 3y(x^2+y^2)-3xy2x }{ (x^2+y^2)^2 }=...=\frac{ 3y(x^2-y^2) }{ (x^2+y^2)^2 }\]\[\frac{ \partial f}{ \partial y }(x, y)=-\frac{ 3x(x^2+y^2)-3xy2y }{ (x^2+y^2)^2 }=...=\frac{ -3x(x^2-y^2) }{ (x^2+y^2)^2 }\]Now I'm going to do some thinking... ;)

Answer 4

@ZeHanz excuse me!! I think you got confuse with my question -.-

Answer 5

\[f _{x}(0,0)=Lt _{h \rightarrow0}f((0+h,0)-f(0,0))/h\]

Answer 6

so........after substitution u get this \[f _{x}(0,0)=0\]

Answer 7

The partial derivatives I calculated are formulas only valid for (x, y) <> (0, 0).
To get the pd's in (0, 0), you only have to calculate:\[\frac{ \partial f }{ \partial x } (0,0)=\lim_{h \rightarrow 0}\frac{ f(0+h, 0)-f(0,0) }{ h }\]and\[\frac{ \partial f }{ \partial y } (0,0)=\lim_{h \rightarrow 0}\frac{ f(0, 0+h)-f(0,0) }{ h }\]The only thing is, with this function f, I don't think that the partial derivatives actually exist in (0,0).

Answer 8

they do exist are equal to zero.........@ZeHanz

Answer 9

u know that f(x,y) is differentiable at the origin(0,0) if \[f(h,k)-f(0,0)-hf _{x}(0,0)-kf _{y}(0,0)=h \epsilon _{1}+k \epsilon _{2}\]

Answer 10

where
\[\epsilon _{1}\] and \[\epsilon _{2}\] tend to zero as h,k tend to zero independently

Answer 11

My fault, the partial derivatives are 0 in (0, 0), I see that now (could use l'H to prove).
If f is discontinuous in (0, 0), then it is not differentiable...

Answer 12

so i have to find f is not continues in (0, 0) then it's both exist right?

Answer 13

@ZeHanz @nitz ........

Answer 14

No, we already know that the partial derivatives exist in (0,0). The both have value 0 there.
We now have to prove that f is not continuous in (0,0). This means that when (x,y) is chosen near to (0,0), f(x,y) is not near to f(0,0)=0.
Because f must at least be continuous in (0,0) to be differentiable there, failing this condition means f is not differentiable in (0,0).

Answer 15

Do you mind telling me more about the solution for this.

Answer 16

Because continuity is a necessary condition for differentiability,I hoped to prove f is not continuous in (0,0). Now I'm not sure f isn't continuous in (0,0). It could well be, because it seems to me the limit of f as (x,y) approaches (0,0) is 0, exactly f(0,0).

If it is continuous, but not differentiable, the graph has some kind of sharp edge in (0,0).
(like in the one-dimensional case the graph of f(x)=|x| has in x=0)

The whole thing is rather complex.
Here is a PowerPoint about it, I've found on the internet:
www2.kenyon.edu/Depts/.../Differentiability.ppt

Answer 17

Even better:
http://higheredbcs.wiley.com/legacy/college/hugheshallett/0471484822/theory/hh_focusontheory_sectioni.pdf

See Example 4 in this document. It is about the function f(x,y)=xy/(x²+y²), so (apart from the non-important factor -3) it's just what you are looking for! Your question is answered there.

Answer 18

It turns out that f is indeed NOT continuous in (0,0), so it is also NOT differentiable, although the partial derivatives exist there...

Answer 19

Why is it not continuous in (0,0)?
To be continuous, this should be true: \[\lim_{(x,y) \rightarrow (0,0)}f(x,y)=f(0,0)=0\]It means, that along whatever path you walk to (0,0), you should always arrive at 0, which is the value of f in (0, 0).

But: if you walk along the path (x,x), i.e. along the line y=x, then the limit becomes:\[\lim_{x \rightarrow 0}\frac{ -3x^2 }{ 2x^2 }=-\frac{ 3 }{ 2 }\neq0\]So it doesn't lead to f(0,0) = 0, but to -3/2.
Conclusion: f not continuous in (0,0), so also not differentiable.

Answer 20

@ZeHanz THank you

Answer 21

YW!