Partial derivative of (x³+y³)^(1/3) at (0,0)

[zhuyilun]

Homework Statement

find the partial derivative of f(x,y)=(x^3+y^3)^(1/3) with respect to x and evaluate at (0,0)

Homework Equations

The Attempt at a Solution

i found the general partial derivative with respect to x is (x^2)*(x^3+y^3)^(-2/3)
if i plug in the point i would get zero at the bottom
so i used the limit thing which is the limit of (f‘(x+h,y)-f(x,y))/h as h approaches infinite.
then i substitute , i got something like lim (((x+h)^3+y^3)^(1/3)-(x^3+y^3)^(1/3))/h as h approaches infinite. then i plug in x=0, y=0, i got lim ((h^3)^(1/3))/h as h approaches infinite which is just 1
i am not sure about what i did is right or not

 

[Avodyne]

Your partial derivative is correct, but the value at (x,y)=(0,0) depends on how you approach this point. It is 1 if you first set y=0 with x positive, and then take the limit as x->0.

Bad question. Complain to your instructor. Seriously.

 

[n!kofeyn]

The partial derivative is simply not defined at (0,0).