Partial derivative of a function at (0,0)

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The discussion centers on calculating the partial derivatives of the function f(x,y) = xy/(x^2+y^2) at the point (0,0). Users note that the function is undefined at this point, leading to confusion when trying to compute the partial derivatives fx and fy. It is emphasized that standard derivative calculations cannot be applied directly at (0,0) due to the function's undefined nature there. To properly find the partial derivatives at this point, one must use the formal definition of partial derivatives, which involves calculating limits. The conversation highlights the importance of recognizing the function's limitations at specific points when performing derivative calculations.
davidp92
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Homework Statement


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So the example says fx(0,0)=0 and fy(0,0)=0 (the partial derivatives).
When I try it I'm getting functions that are not defined at (0,0):
f(x,y)=xy/(x^{}+y^{})
so for example,
fx=[x(x^2+y^2)-2y(xy)]/(x^2+y^2)^2
fx=(x^3+xy^2-2xy^2)/(x^2+y^2)^2
fx=x^3-xy^2/(x^2+y^2)^2

Which I keep getting fx(0,0) being undefined. What am I doing wrong?
 
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Your getting this because the expression for f that you use, is undefined in 0. That is, when you calculate the partial derivatives, then it's ok to derive the form f(x,y)=\frac{xy}{x^2+y^2} for every point except (0,0). But this will not help us in (0,0).

To find the partial derivatives in (0,0), you'll going to have to use the definition, I'm afraid. What is the definition of the partial derivative?? Can you calculate the limit involved?
 
I recognize that typesetting anywhere. It's mr.stweart's hideous textbook. I am exploring more of this topic, but vela said that you don't even assume it is defined at the origin in the first place
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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