Differentiablity of $\cos|xy|$ at $(0,0)$ Is $\cos|xy|$ differentiable at $(0,0)$?
I've tried to work it out by definition but I couldn't compute the limit
$$\lim_{(x,y)\to(0,0)}{\cos|xy|-1\over \sqrt{x^2+y^2}}$$
 A: Yes.
Even though the absolute value appears problematic, the cosine function is even, making the absolute value go away. In other words $\cos |xy|=\cos (\pm xy)=\cos xy$ which is definitely differentiable at (0,0) (and at any other point).
A: If we use the MacLaurin series expansion of cosine, we see that $$\cos|xy|-1=\sum_{n=1}^\infty\frac{(-1)^n|xy|^{2n}}{(2n)!}=-|xy|^2\sum_{n=0}^\infty\frac{(-1)^n|xy|^{2n}}{(2n+2)!}=-x^2y^2\sum_{n=0}^\infty\frac{(-1)^n(x^2y^2)^n}{(2n+2)!},$$ and it can be seen readily by continuity that $$\lim_{(x,y)\to(0,0)}\sum_{n=0}^\infty\frac{(-1)^n(x^2y^2)^n}{(2n+2)!}=\frac12.$$ Hence, $$\lim_{(x,y)\to(0,0)}\frac{\cos|xy|-1}{\sqrt{x^2+y^2}}=-\frac12\lim_{(x,y)\to(0,0)}\frac{x^2y^2}{\sqrt{x^2+y^2}},$$  I leave it to you to evaluate the limit on the right.

Another way to see this is to observe that cosine is an even function, so $\cos|xy|=\cos(-xy)=\cos(xy).$ Since $(x,y)\mapsto xy$ is a differentiable function $\Bbb R^2\to\Bbb R$ and cosine is differentiable as a real-valued function on $\Bbb R$, then $\cos|xy|$ is differentiable.
A: This might be easier in polar coordinates:
$x=r\cos \left( \theta \right)$ 
$y=r\sin \left( \theta \right)$ 
${\frac {\cos \left( xy \right) -1}{\sqrt {{x}^{2}+{y}^{2}}}}={\frac {
\cos \left( 1/2\,{r}^{2}\sin \left( 2\,\theta \right)  \right) -1}{r}}$ 
Taylor expand in $r$ near the origin:
${\frac {
\cos \left( 1/2\,{r}^{2}\sin \left( 2\,\theta \right)  \right)-1 }{r}}=  -1/8\, \left( \sin \left( 2\,\theta
 \right)  \right) ^{2}{r}^{3}+{\frac {1}{384}}\, \left( \sin \left( 2
\,\theta \right)  \right) ^{4}{r}^{7}+O \left( {r}^{11}  
 \right)$ 
So as $r$ goes to zero you have, for all angles: 
$\lim_{r\rightarrow 0}  \left( {\frac {
\cos \left( 1/2\,{r}^{2}\sin \left( 2\,\theta \right)  \right) -1}{r}}
 \right)=0
$
