Let f(x,y)=2xy/(x^2+y^2)^1/2
(x,y) not equal to (0,0) , f(0,0)=0
Show fyx(0,0) not equal to fxy(0,0)
using method fxy= lim k->0 [ fx(0,k)-fx(0,0)]/k
fyx=lim h->0 [ fy(h,0)-fy(0,0)]/h

- anonymous

- jamiebookeater

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- apoorvk

umm, this may be stupid, but how can the function be defined for (0,0) when it already says (x,y) not equal to (0,0). [read second line of the problem]
am i missing something?

- anonymous

Well nothing is stupid . Thats how you prove limits. Google it !

- apoorvk

yeah i know limits, but then you must mean in the second line that
lim f(0,0) = 0
x,y->0
right??

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## More answers

- anonymous

yup !

- anonymous

What does fx(0,k) mean? x->0 and y -> k?

- anonymous

But f(0,0) isn't defined, is it?

- anonymous

Maybe @dumbcow knows how to do it, this stuff is a little advanced for me. I'm fairly certain this isn't a complicated problem, but I haven't taken multi-variable calculus yet.

- apoorvk

okay so i seem to think fxy here is partial differential of f(x,y) wrt to x, and fyx is partial differential of f(x,y) wrt to y. see the structure. and same here, haven't taken advanced calculus, so don't think i ll be able help much other than guessing.

- dumbcow

interesting, i find that fxy = fyx , im assumimg these are second partial derivatives
\[\large f_{xy} = f_{yx} = \frac{6x^{2}y^{2}}{(x^{2}+y^{2})^{5/2}}\]
http://www.wolframalpha.com/input/?i=d%2Fdx+d%2Fdy+2xy%2Fsqrt%28x^2%2By^2%29
http://www.wolframalpha.com/input/?i=d%2Fdy+d%2Fdx+2xy%2Fsqrt%28x^2%2By^2%29

- dumbcow

since they are equavalent, the limit at (0,0) would also be the same

- apoorvk

yeah i logically was also getting that conclusion, but ofcourse more based on logical guessing and less of calculations.

- dumbcow

maybe the question was show they are equal ?

- apoorvk

may be yes. high chances of that.

- anonymous

The question is that show they are not equal. I stated the method to be used , and that's the way I have to do it in my exam , and not just simple estimation.

- anonymous

Actually its a "step by step" sort of way , but I was not able to prove the conclusion , maybe some logical error Iam facing.

- anonymous

lets try this
first off we need \(f_x(x,y)=\frac{2y^3}{(x^2+y^2)^{\frac{3}{2}}}\)

- anonymous

and so \(f_x(0,y)=\frac{2y^3}{(y^2)^{\frac{3}{2}}}=2\) if i am not mistaken

- anonymous

well hold on this is symmetric in x and y, so why these would not be the same is now confusing me. hmmm

- anonymous

Read the following link. Complete the steps, then mimic for your function
http://www.unf.edu/~omilatov/MAC2313/differentiability.pdf

- anonymous

@ satellite , the method Iam talking about , you are not following that , maybe thats what misleading you !

- anonymous

@elias, I know the steps , its that Iam not able to prove the inequality , some logical error that too in the last steps are over shadowing the ans ! So see if you could solve it step by step

- Zarkon

I would check to make sure that you wrote down/typed the correct function

- anonymous

yup the function is correct , i copied directly from my book .

- anonymous

and its sort of a proof , and not just straightforward estimation of fxy or fyx , maybe thats why many are getting confused !

- anonymous

Those two limits seem to be both infinite.

- anonymous

\[ f_x(x,y)=\frac{2
y^3}{\left(x^2+y^2\right)^{3/
2}}\\
f_y(x,y)=\frac{2
x^3}{\left(x^2+y^2\right)^{
3/2}}
\]
You can show from the definition that
\[ f_x(0,0)= f_y(0,0)=0\]
so
\[\frac{ f_x(0,k) - 0}{k}=\frac{2}{\sqrt{k^2}}
\] and
\[\frac{ f_y(k,0) - 0}{k}=\frac{2}{\sqrt{k^2}}
\]

- anonymous

So the limit when x tends to zero is infinite for both.

- anonymous

yea even I was getting the same ans ! but since the ques demands something different , I got stuck in between ! Then I thought since its coming to be infinity , so it must be implying that limits does not exist !

- anonymous

yes.

- anonymous

pardon ?

- anonymous

Do not worry about it.

- anonymous

From May its my papers , so have a reason to worry ! :-(

- anonymous

i think there is something wrong here because this function is symmetric in x and y, i.e. if you replace x by y and y by x you get the same thing. so why one should be different from the other is not at all clear to me

- anonymous

:-(((((( no idea ! :-(((((

- anonymous

Both limits diverge to \[ \infty \]
In a way they are not different.

- anonymous

They are not different , but does not exist either !

- anonymous

Yes.

- anonymous

Iam gettting your point , infact I myself concluded the same ans , but ques says something else. I ll confirm it tomorrow and will let you know. Thanks a lot for your help ! :-)

- apoorvk

phew... please don't forget to confirm what this was all about!

- anonymous

lol ! Yea this method is a bit "mugging up" thing , maybe you have never read that, so you were getting confused so as to what to do ! Anyways google it ,its a proper method ! And yea thanks ! :-)

- apoorvk

definitely mugging up method i believe. atleast very theoretical or subjective. bet it's for a theory paper in your uni, right?

- anonymous

actually maths is a subsi in my course ! But yea it must be there in maths(h) ! IMP PROOOF ! :P

- anonymous

Please read http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives
The above problem is a bit related to the link above

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