Partial Derivatives Problem Evaluating at (0,0)

In summary, the problem involves evaluating the derivatives at (0,0) for a given function using Matlab. The program gave the same answer for both fxy and fyx, which is incorrect according to the problem. The solution involves calculating the derivatives using the definition of the limit and simplifying the expressions to get the correct values for fxy and fyx at (0,0).
  • #1
Jamie2
17
0
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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  • #2
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 \(\displaystyle f_x(0,0) = \lim_{h\to0}\frac{f(h,0) - f(0,0)}h = \lim_{h\to0}\frac{\frac {h*0(h^2-0^2)}{h^2+0^2} - 0}h = 0.\) The next step is to calculate \(\displaystyle f_{xy}(0,0) = \frac{\partial f_x}{\partial y}(0,0) = \lim_{k\to0}\frac{f_x(0,k) - f_x(0,0)}k\) in a similar way. Then do the same for $f_y(0,0)$ and $f_{yx}(0,0)$.
 
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  • #3
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 \(\displaystyle f_x(0,0) = \lim_{h\to0}\frac{f(h,0) - f(0,0)}h = \lim_{h\to0}\frac{\frac {h*0(h^2-0^2)}{h^2+0^2} - f(0,0)}h = 0.\) The next step is to calculate \(\displaystyle f_{xy}(0,0) = \frac{\partial f_x}{\partial y}(0,0) = \lim_{h\to0}\frac{f_x(0,k) - f_x(0,0)}k\) 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?
 
  • #4
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 \(\displaystyle f_{xy}(0,0) = \lim_{k\to0}\frac{f_x(0,k) - f_x(0,0)}k = \lim_{k\to0}\frac{-k - 0}k = -1.\) My guess is that when you do the same thing for $f_{yx}(0,0)$, the answer will come out as $+1.$
 
  • #5
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 \(\displaystyle f_{xy}(0,0) = \lim_{k\to0}\frac{f_x(0,k) - f_x(0,0)}k = \lim_{k\to0}\frac{-k - 0}k = -1.\) 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.
 
  • #6
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)
 

FAQ: Partial Derivatives Problem Evaluating at (0,0)

What are partial derivatives?

Partial derivatives are a type of derivative used in multivariable calculus to calculate the rate of change of a function with respect to one of its variables while holding all other variables constant. They allow us to analyze how a function changes in different directions.

How do you evaluate a partial derivative at a specific point?

To evaluate a partial derivative at a specific point, you need to plug in the given values for all the variables in the function and then solve the resulting equation. This will give you the value of the partial derivative at that point.

What is the significance of evaluating a partial derivative at (0,0)?

Evaluating a partial derivative at (0,0) can reveal important information about the behavior of the function at the origin. It can help us determine if the function is continuous or differentiable at that point, and if it has any critical points or extrema.

How is the process of evaluating a partial derivative at (0,0) different from evaluating a regular derivative?

The process of evaluating a partial derivative at (0,0) is similar to evaluating a regular derivative, but with an added step. In addition to plugging in the given values for the variables, we also have to hold all other variables constant. This means that we treat them as constants and do not include them in our final answer.

In what real-life applications are partial derivatives used?

Partial derivatives have many practical applications, such as in economics, physics, engineering, and statistics. They are used to analyze the behavior of complex systems and make predictions about how they will change. For example, they can be used to optimize production processes, model the growth of populations, and determine the rate of change of temperature in a room.

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