Differentiability of $xy^{\alpha}$ I was asked to prove that
$|xy|^{\alpha}$ is differentiable at $(0,0)$ if $\alpha > \frac{1}{2}$.
Since both the partial derivatives are zero, I concluded that this function is differentiable if and only if the following holds:
$$ \lim\limits_{(x,y)\to (0,0)} \frac{|xy|^{\alpha}}{\sqrt{x^{2} + y^{2}}} = 0$$
However, I am not sure how to show this. What I tried is:
$$\frac{|xy|}{\sqrt{x^{2}+y^{2}}} \leq \frac{1}{\sqrt{2}}$$
Hence the given expression is less than equal to
$$ \frac{1}{\sqrt{2}}|xy|^{\alpha - \frac{1}{2}}$$
Now I conclude that this goes to zero as $x,y$ go to zero?
I was just wondering if I am correct in all my steps?
Any help would be apreciated.
EDIT: Can I use other norms on $\mathbb{R}^{2}$ instead of the Euclidean norm to conclude matters of differentiability? I ask this as it would be easier to work with other norms.

Polar co-ordinates. 
$\displaystyle \lim_{(x,y)\to(0,0)}\frac{|xy|^{\alpha}}{\sqrt{x^2+y^2}}=\lim_{r\to 0}r^{2\alpha-1}|\sin \theta \cos \theta|$ which $\to 0$ if $\alpha>\frac{1}2$
