Partial Derivatives Problem Evaluating at (0,0)

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SUMMARY

The discussion centers on evaluating partial derivatives of the function f(x,y) = xy(x^2 - y^2)/(x^2 + y^2) at the point (0,0) using MATLAB and the limit definition of derivatives. Users encountered issues with MATLAB returning the same values for fxy and fyx, which is incorrect at (0,0). The correct approach involves calculating the limits for fxy(0,0) and fyx(0,0) from first principles, leading to the conclusion that fxy(0,0) = -1 and fyx(0,0) = +1.

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Jamie2
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Problem: View attachment 2084I did some of the problem on MatLab but I'm having a difficult time evaluating the derivatives at (0,0). Also, MatLab gave me the same answer for fxy and fyx, which according to the problem isn't correct. Any ideas? I used MatLab and computed:

fx(x,y)=(2*x^2*y)/(x^2 + y^2) + (y*(x^2 - y^2))/(x^2 + y^2) - (2*x^2*y*(x^2 - y^2))/(x^2 + y^2)^2

and

fy(x,y)=(x*(x^2 - y^2))/(x^2 + y^2) - (2*x*y^2)/(x^2 + y^2) - (2*x*y^2*(x^2 - y^2))/(x^2 + y^2)^2

I also used MatLab to compute fxy and fyx, both gave me the same answer:

(x^2 - y^2)/(x^2 + y^2) + (2*x^2)/(x^2 + y^2) - (2*y^2)/(x^2 + y^2) - (2*x^2*(x^2 - y^2))/(x^2 + y^2)^2 - (2*y^2*(x^2 - y^2))/(x^2 + y^2)^2 + (8*x^2*y^2*(x^2 - y^2))/(x^2 + y^2)^3
 

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The Matlab answers for $f_{xy}$ and $f_{yx}$ are valid when $(x,y)\ne(0,0)$. The only point at which $f_{xy} \ne f_{yx}$ is when $(x,y) = (0,0).$ So you need to calculate $f_{xy}(0,0)$ and $f_{yx}(0,0).$ You need to do this from first principles, using the definition of the derivative as a limit.

You first need to check that $$f_x(0,0) = \lim_{h\to0}\frac{f(h,0) - f(0,0)}h = \lim_{h\to0}\frac{\frac {h*0(h^2-0^2)}{h^2+0^2} - 0}h = 0.$$ The next step is to calculate $$f_{xy}(0,0) = \frac{\partial f_x}{\partial y}(0,0) = \lim_{k\to0}\frac{f_x(0,k) - f_x(0,0)}k$$ in a similar way. Then do the same for $f_y(0,0)$ and $f_{yx}(0,0)$.
 
Last edited:
Opalg said:
The Matlab answers for $f_{xy}$ and $f_{yx}$ are valid when $(x,y)\ne(0,0)$. The only point at which $f_{xy} \ne f_{yx}$ is when $(x,y) = (0,0).$ So you need to calculate $f_{xy}(0,0)$ and $f_{yx}(0,0).$ You need to do this from first principles, using the definition of the derivative as a limit.

You first need to check that $$f_x(0,0) = \lim_{h\to0}\frac{f(h,0) - f(0,0)}h = \lim_{h\to0}\frac{\frac {h*0(h^2-0^2)}{h^2+0^2} - f(0,0)}h = 0.$$ The next step is to calculate $$f_{xy}(0,0) = \frac{\partial f_x}{\partial y}(0,0) = \lim_{h\to0}\frac{f_x(0,k) - f_x(0,0)}k$$ in a similar way. Then do the same for $f_y(0,0)$ and $f_{yx}(0,0)$.
the problem is that when I try to use the limit definition of the derivative I get that it's undefined (0/0). Do you have any suggestions for how I can compute that limit?
 
Jamie said:
the problem is that when I try to use the limit definition of the derivative I get that it's undefined (0/0). Do you have any suggestions for how I can compute that limit?
The result I get by using the quotient rule (which I hope agrees with the Matlab result except that I have tried to simplify it a bit) is $$f_x(x,y) = \frac{\bigl(y(x^2-y^2) + 2x^2y\bigr)(x^2+y^2) - 2x^2y(x^2-y^2)}{(x^2+y^2)^2} = \frac{y(x^4-y^4) + 4x^2y^3}{(x^2+y^2)^2}.$$ Then when you put $x=0$ and $y=k$ you get $$f_x(0,k) = \frac{-k^5}{k^4} = -k.$$ Now put that into the expression for $f_{xy}(0,0)$ and you get $$f_{xy}(0,0) = \lim_{k\to0}\frac{f_x(0,k) - f_x(0,0)}k = \lim_{k\to0}\frac{-k - 0}k = -1.$$ My guess is that when you do the same thing for $f_{yx}(0,0)$, the answer will come out as $+1.$
 
Opalg said:
The result I get by using the quotient rule (which I hope agrees with the Matlab result except that I have tried to simplify it a bit) is $$f_x(x,y) = \frac{\bigl(y(x^2-y^2) + 2x^2y\bigr)(x^2+y^2) - 2x^2y(x^2-y^2)}{(x^2+y^2)^2} = \frac{y(x^4-y^4) + 4x^2y^3}{(x^2+y^2)^2}.$$ Then when you put $x=0$ and $y=k$ you get $$f_x(0,k) = \frac{-k^5}{k^4} = -k.$$ Now put that into the expression for $f_{xy}(0,0)$ and you get $$f_{xy}(0,0) = \lim_{k\to0}\frac{f_x(0,k) - f_x(0,0)}k = \lim_{k\to0}\frac{-k - 0}k = -1.$$ My guess is that when you do the same thing for $f_{yx}(0,0)$, the answer will come out as $+1.$

I am not sure that this is what the question is asking. Basically I just need help solving the limit definition of derivative algebraically because every time I try I get 0/0= undefined.
I need to show the value of the derivative for f(x,y)= xy(x^2-y^2)/(x^2+y^2), at fx(0,0) and at fy(0,0) using the limit definition of derivative for both.
 
Jamie said:
I am not sure that this is what the question is asking. Basically I just need help solving the limit definition of derivative algebraically because every time I try I get 0/0= undefined.
I need to show the value of the derivative for f(x,y)= xy(x^2-y^2)/(x^2+y^2), at fx(0,0) and at fy(0,0) using the limit definition of derivative for both.
If you look at part (iii) of the question, you will see that that is exactly what I have indicated how to do. (Nod)
 

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