- #1
Joker93
- 504
- 37
Homework Statement
I have the function:
f(x,y)=x-y+2x^3/(x^2+y^2) when (x,y) is not equal to (0,0). Otherwise, f(x,y)=0.
I need to find the partial derivatives at (0,0).
With the use of the definition of the partial derivative as a limit, I get df/dx(0,0)=3 and df/dy(0,0)=-1. However, my problem here is that if I just compute the derivative in the standard way and then take the limit (x,y)-->(0,0) I get that the derivatives don't exist at the origin.
Homework Equations
df/dx=1+(2x^4+6x^2*y^2)/((x^2+y^2)^2
df/dy=-1-4x^3*y)/(x^2+y^2)^2
The Attempt at a Solution
When I take the limit (x,y)->(0,0) and I put y=mx I get:
df/dx=1+(2+6m^2)/(1+m^2)^2
df/dy=-1-(4m)/(1+m^2)^2
So, clearly both limits depend on the value of m, so they do not exist. So, why does the original method(using the definition of the derivative) wield the values 3 and -1 for each respectively?
I also checked the answer to the problem and the answer gives 3 and -1 and quotes "by using the definition of the partial derivatives".
Thanks in advance.