# Thread: Show function is differentiable over its domain

1. ## Show function is differentiable over its domain

I need to show that:

$\displaystyle F(x,y,z) = \frac{x + y + z}{x^2+y^2+z^2}$

is differentiable over its domain. The only point I see there being a problem at is when (x,y,z) = (0,0,0).

Then is it sufficient to show that:

$\displaystyle \lim_{(x,y,z)\to (0,0,0)}\frac{f(x,y,z) - h(x,y,z)}{|| (x,y,z) - (0,0,0) ||} = 0$

where $\displaystyle h(x,y,z) = f(0,0,0) + f_x(0,0,0)(x) +$
$\displaystyle f_y(0,0,0)(y) + f_z(0,0,0)(z)$ for this case.

I do see a problem in the h(x,y,z) function regarding f(0,0,0). That's what I am trying to find the limit of, and in my function I'd get 0/0 when I plug in those values. Also, I am getting the partial derivatives to be of the form:
$\displaystyle f_x(x,y,z) = \frac{x^2+y^2+z^2 - (x + y + z)2x}{(x^2+y^2+z^2)^2}$
$\displaystyle f_y(x,y,z) = \frac{x^2+y^2+z^2 - (x + y + z)2y}{(x^2+y^2+z^2)^2}$
$\displaystyle f_x(x,y,z) = \frac{x^2+y^2+z^2 - (x + y + z)2y}{(x^2+y^2+z^2)^2}$

Using all the info I seem to get many terms who wind up becoming 0/0. I believe I am missing the point and/or missing a way to simplify everything. Thanks for any help.

2. I just realized that the functions domain is:

all R3 except for (0,0,0).

The function isn't defined at that point and so it isn't in its domain.

Does that mean I can simply use the above formula to show that for all (a,b,c) the limit exists? If that's the case then I suppose I would need to show that the limit = 0 for all (a, b, c) correct? The reason is that for anyway I look at the limit at (0,0,0) I appear to be getting that it doesn't exist.