Find if $\sqrt[4]{x^4+y^4}, \sqrt{x^4+y^4}$ are differentiable in $(0,0)$ 
Find if
  $$f(x,y)=\sqrt[4]{x^4+y^4}$$
  $$g(x,y)=(f(x,y))^2$$
  are differentiable in $(0,0)$.

well, $g(x)$ is clearly $\sqrt{x^4+y^4}$, so I guess the answer will be similar to $f(x)$.
$f_x=x^3/(x^4+y^4)^{3/4}$, but what now? $f_x$ has no value in $(0,0)$, so can it be differentiable there?
 A: HINT: 
The sufficient conditions for differentiability of a function $f(x,y)$ at $(a,b)$ are-


*

*The partial derivatives of $f$ i.e. $f_x$ and $f_y$ must exist at $(a,b)$.

*The partial derivatives of $f$ must be continuous at $(a,b)$.

A: Using the definition: in both cases, as $(0,0)$ is a local minimum, the (possible) differential will be zero if it exists:
$$
\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-(0,0)(x,y)}{\|(x,y)\|} =
\lim_{(x,y)\to(0,0)}\frac{\sqrt[4]{x^4+y^4}}{\sqrt{x^2+y^2}} =
\lim_{r\to 0}\frac{r(\cos^4\theta+\sin^4\theta)}r.
$$
($\not\exists$ because depends of the angle)
$$
\lim_{(x,y)\to(0,0)}\frac{g(x,y)-g(0,0)-(0,0)(x,y)}{\|(x,y)\|} =
\lim_{(x,y)\to(0,0)}\frac{\sqrt{x^4+y^4}}{\sqrt{x^2+y^2}} =
\lim_{r\to 0}\frac{r^2(\cos^4\theta+\sin^4\theta)}r = 0.
$$
And only $g$ is differentiable.
EDIT: clarification of second limit:
$$
\left|\frac{\sqrt{x^4+y^4}}{\sqrt{x^2+y^2}}\right| =
\frac{r^2(\cos^4\theta+\sin^4\theta)}r \le 2 r = 2\|(x,y)\|\to 0,
$$
and the squeezing theorem proves that the lim is 0.
