Here's the question you clicked on:
anjali_pant
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
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?
Well nothing is stupid . Thats how you prove limits. Google it !
yeah i know limits, but then you must mean in the second line that lim f(0,0) = 0 x,y->0 right??
What does fx(0,k) mean? x->0 and y -> k?
But f(0,0) isn't defined, is it?
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.
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.
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
since they are equavalent, the limit at (0,0) would also be the same
yeah i logically was also getting that conclusion, but ofcourse more based on logical guessing and less of calculations.
maybe the question was show they are equal ?
may be yes. high chances of that.
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.
Actually its a "step by step" sort of way , but I was not able to prove the conclusion , maybe some logical error Iam facing.
lets try this first off we need \(f_x(x,y)=\frac{2y^3}{(x^2+y^2)^{\frac{3}{2}}}\)
and so \(f_x(0,y)=\frac{2y^3}{(y^2)^{\frac{3}{2}}}=2\) if i am not mistaken
well hold on this is symmetric in x and y, so why these would not be the same is now confusing me. hmmm
Read the following link. Complete the steps, then mimic for your function http://www.unf.edu/~omilatov/MAC2313/differentiability.pdf
@ satellite , the method Iam talking about , you are not following that , maybe thats what misleading you !
@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
I would check to make sure that you wrote down/typed the correct function
yup the function is correct , i copied directly from my book .
and its sort of a proof , and not just straightforward estimation of fxy or fyx , maybe thats why many are getting confused !
Those two limits seem to be both infinite.
\[ f_x(x,y)=\frac{2 y^3}{\left(x^2+y^2\right)^{3/ 2}}\\ f_y(x,y)=\frac{2 x^3}{\left(x^2+y^2\right)^{ 3/2}} \] 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 \[\frac{ f_y(k,0) - 0}{k}=\frac{2}{\sqrt{k^2}} \]
So the limit when x tends to zero is infinite for both.
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 !
Do not worry about it.
From May its my papers , so have a reason to worry ! :-(
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
:-(((((( no idea ! :-(((((
Both limits diverge to \[ \infty \] In a way they are not different.
They are not different , but does not exist either !
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 ! :-)
phew... please don't forget to confirm what this was all about!
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 ! :-)
definitely mugging up method i believe. atleast very theoretical or subjective. bet it's for a theory paper in your uni, right?
actually maths is a subsi in my course ! But yea it must be there in maths(h) ! IMP PROOOF ! :P
Please read http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives The above problem is a bit related to the link above