Differentiability of absolute value

[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 doesn't go to 0 it seems. Any help is appreciated thanks!

 

[lippyka]

Yes, you are right that you need to look at the cases separately. For y>0 and h2>0, we have f(x,y) = x√y, so the partial derivative with respect to y is 1/2 x/√y. At (0, 0), this derivative is 0/0, which is indeterminate. To show that the limit exists, we can use the following limit definition:lim |h| ->0 (f((0,0) + (h1,h2)) - f(0,0)) / |h|= lim |h| ->0 (h1* √h2) / |h|= lim |h| ->0 h1 * (√h2 / |h|)= 0 * lim |h| ->0 (√h2 / |h|) = 0 Therefore, at (0, 0), f is differentiable with a Jacobian of [0 0].