MHB Partial Derivatives Problem Evaluating at (0,0)

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The discussion centers on evaluating partial derivatives at the point (0,0) for the function f(x,y) = xy(x^2 - y^2)/(x^2 + y^2). Users are encountering difficulties with the limit definition of derivatives, specifically obtaining undefined results (0/0) when trying to compute fxy and fyx at (0,0). It is emphasized that the correct approach involves using first principles to evaluate these limits directly. The calculations suggest that fxy(0,0) results in -1, while fyx(0,0) is likely +1, indicating that the mixed partial derivatives are not equal at this point. The discussion concludes with a focus on the need for algebraic manipulation to resolve the limits effectively.
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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