Show $f$ is differentiable and find the derivative at any given point.

Asked Feb 12 '14 at 22:43

Active Feb 12 '14 at 22:43

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$f:U \to \mathbb{R}, (x,y) \mapsto \sqrt{1-x^2-y^2}, \text{ where } U = \{(x,y)|x^2+y^2<1\}$

Okay, basically my strategy was: assume that partials exist then do them out and,

Continuous partials $\implies$ differentiable $\implies$ partials exist.

Then I got: $\frac{\partial f}{\partial x} = -x(1-x^2-y^2)^{-1/2}$, $\frac{\partial f}{\partial y} = -y(1-x^2-y^2)^{-1/2}$.

Now I am stuck, because I don't know anything about this $U$. Can it contain complex pairs? Can I assume that $U \subset \mathbb{R}^2$?

Or, if my strategy is wrong, what is the correct one?

asked Feb 12 '14 at 22:43

Bobby Lee

You just can say that on $U$ the partial derivatives are well-defined and continuous as you can check from their expressions. Hence, $f$ is differentiable at any point in $U$ (observe $U$ is open and the critical points which lie on the boundary are excluded) – Martingalo Feb 12 '14 at 22:46
1

$U$ is a set, not a point, so $U\in\mathbb R^2$ makes no sense. – Andrés E. Caicedo Feb 12 '14 at 22:47
Yes, $U$ is the unit ball in $\Bbb R^2$. – Jonathan Y. Feb 12 '14 at 22:48
I meant subset, it was a typo. Anyways, how do I know that the partial derivatives are well-defined? – Bobby Lee Feb 12 '14 at 22:50
Cause the only problem would be having zero or negative values in the square root which can not happen for values in $U$ – Martingalo Feb 12 '14 at 22:51
Makes sense. Thanks. – Bobby Lee Feb 12 '14 at 22:53

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