Question

Let f(x,y) = (x^2 +y^2)^(2/3). Show that

fx(x,y) = {4x/((3)(x^2+y^2)^1/3), (x,y) does not
equal (0,0)
and
0, (x,y)=(0,0)

Answer 1

@zepdrix help?

Answer 2

I can do the first part, but not the second part.

Answer 3

this is partial differentiation again, thanks for helping.

Answer 4

So you were able to find \(\large f_x(x,y)\) without any trouble?
I don't understand what the next part is saying D: hmm

Answer 5

yes, but for the next part, I need to show that fx is 0 when x=0 and y=0.

Answer 6

If i differentiate it as usual, i'll end up with (x^2 +y^2)^1/3 at the bottom, and a denominator can't be zero, so...

Answer 7

Oh hmmm D: Is there some type of limiting process we can do? Hmmmm

Answer 8

can we have
\[\lim_{(x,y) \rightarrow (0,0)} \frac{ 4x }{ 3(x^2+y^2)^\frac{ 1 }{ 3 } }\]

Answer 9

since (x,y) can't be (0,0), instead of substituting it, we can limit it to (0,0)? I'm not sure.

Answer 10

Hmm yah maybe that would work.
I was thinking maybe we use the limit definition of the derivative.
But I dunno :d
\[\Large f_x(x,y)=\lim_{h\to0}\frac{\left[(x+h)^2+y^2\right]^{2/3}-\left[x^2+y^2\right]^{2/3}}{h}\]

\[\Large f_x(0,0)=\lim_{h\to0}\frac{\left[(0+h)^2+0^2\right]^{2/3}-\left[0^2+0^2\right]^{2/3}}{h}\]

Answer 11

yeah, that works, nice, thanks!

Answer 12

Oh cool! ^^