Integration of absolute value

1. Nov 18, 2012

samithie

f: R2 to R1 given by f(x,y) = x(|y|^(1/2))
show differentiable at (0,0)

so I'm using the definition lim |h| ->0 (f((0,0) + 9(h1,h2)) - f(0,0) - Df(0,0) (h1,h2)) / |h|

so first for the jacobian for f, when I'm doing the partial with respect to y, do I have to break this into the case y>0 and y<0 and show its differentiable in both cases (and maybe also have to do the same for when h2 >0 or <0) or can I use do it in one step by rewriting and by differentiating (y^2)^1/4 and. I did that and the Df(0,0) just goes away and then limit doesnt go to 0 it seems. Any help is appreciated thanks!

2. Nov 18, 2012

SammyS

Staff Emeritus

It looks like you have some typo's in your definition (of whatever, you don't say),
lim |h| ->0 (f((0,0) + 9(h1,h2)) - f(0,0) - Df(0,0) (h1,h2)) / |h|​

Clearly, the partial derivative, ∂f(x,y)/∂y, does not exist for y=0, except perhaps when x=0 and y=0 . Of course, that's at least part the matter at hand ...

I'm no expert on this, but I suggest looking at the directional derivative of f(x,y) at (0,0), for arbitrary direction.

3. Nov 19, 2012

samithie

yes sorry about the typos. I think the directional derivatives is not the issue I'm just wondering about if I need to do this piecewise.

4. Nov 19, 2012

SammyS

Staff Emeritus
√(|y|) is equivalent to (y2)1/4