Partial derivative of a function at (0,0)

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 6K views
davidp92
Messages
12
Reaction score
0

Homework Statement


tu6DA.jpg

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[itex]^{}[/itex]+y[itex]^{}[/itex])
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?
 
Physics news on Phys.org
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 [itex]f(x,y)=\frac{xy}{x^2+y^2}[/itex] 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