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 Exercise [10.5] 
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Joined: 05 Jan 2009, 15:24
Posts: 4
Post Exercise [10.5]
I have tried to do this problem by brute force (just bloody calculating the mixed derivatives) and I get that both partial derivatives are equal at the origin. I find that \begin{align*}<br />\frac{\partial}{\partial x}\left ( \frac{\partial f(x,y)}{\partial y}\right)=&\frac{8 x^2 y^2 \left(x^2-y^2\right)}{\left(x^2+y^2\right)^3}-\frac{2 x^2 \left(x^2-y^2\right)}{\left(x^2+y^2\right)^2}-\frac{2 y^2 \left(x^2-y^2\right)}{\left(x^2+y^2\right)^2}+\frac{2 x^2}{x^2+y^2}-\frac{2 y^2}{x^2+y^2}+\frac{x^2-y^2}{x^2+y^2}\\<br />=&\frac{\partial}{\partial y}\left ( \frac{\partial f(x,y)}{\partial x}\right)<br />\end{align*}

(Incase that didn't come out, I have also attached a pdf.)

What worries me is that I also used Mathematica to check this, and the result is still the same irrespective of the order of differentiation.

So, does anybody know an easier method of doing this, and what is going wrong? Does it have something to do with limits, as we need to compare at the origin? Thanks!

temp.pdf [15.63 KiB]
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Last edited by Runner on 24 Oct 2009, 18:33, edited 1 time in total.
05 Jan 2009, 15:49

Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: Excercise [10.5]
Yes, the expression is the same, but you cannot then argue that the derivatives are equal at the origin since if you substitute x=0,y=0 you get expressions like 0/0. You need to look at what's happening near the origin.

For example:
Putting x=0 and y not equal to 0, your derivative =-1 (y not equal 0)
Putting y=0 and x not equal to 0, your derivative =+1 (x not equal 0)
So coming towards the origin along the y axis your derivative is -1
and coming towards the origin along the x-axis your derivative is +1
Approaching along the lines y=x and y=-x your derivative=0

I suggest you do Ex [10.03] before doing this exercise.

05 Jan 2009, 19:22
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