Let f(x,y) =(x^3 +y^3) ^(1/3)
At what points if any, does fy(x,y) fail to exist?

Question

Let f(x,y) =(x^3 +y^3) ^(1/3)
At what points if any, does fy(x,y) fail to exist?

anonymous · Answer

$f_y(x,y)=y^2(x^3+y^3)^{-2/3}$
so it will not exist where $x^3+y^3=0$, or to write it other way $y^3=-x^3$
solve this equation to get the points where this happens

anonymous · Answer

good work @myko

tiffanymak1996 · Answer

in the answer of my book,
equation 
$$y \neq0$$
is also an answer along with y=-x, how do i get y does not equal 0?

anonymous · Answer

Other way to write it:
y^3+x^3=(x+y)(x^2-xy+y^2)=0
so you see that x+y=0 makes the hole thing to be 0, right? so y=-x is a place where $f_y$ will not exist, and it's a line trough the origin

anonymous · Answer

ty @oldrin.bataku

tiffanymak1996 · Answer

well, i can also work that out from y^3 = -x^3 , but what about y does not equal 0?

anonymous · Answer

$y \neq 0$ by it self is not a place where $f_y$ not exist.

anonymous · Answer

maybe what your book means is that if $y \neq 0$ then it have to be on the y=-x line

tiffanymak1996 · Answer

oh, ok, thx!

anonymous · Answer

http://www.wolframalpha.com/input/?i=d%5Bcbrt%28x%5E3%2By%5E3%29%5D%2Fdy