Show differentiability of $f(x) = \sqrt{x^4 + y^4}$ in $(0,0)$

Question

I'm trying to show the differentiability of $$f:\mathbb R^2 \to \mathbb R\text;\quad f(x) = \sqrt{x^4 + y^4}$$ in (0,0). Here's my attempt:
Since $\partial_xf(x,y) = \frac{4x^3}{2\sqrt{x^4+y^4}}$ we would divide be zero plugging in (0,0), so we need to compute the limit manually (same holds for $\partial_y f(x,y)$). Doing that, we get $$\lim \limits_{h\to 0} \frac{f(h,0) - f(0)}{h} = \lim \limits_{h\to 0} \frac{h^2}{h} = \lim \limits_{h\to 0} h = 0$$ (analog for $\partial_y f(x,y)$), hence $\partial_xf(0,0) = 0$. Now it remains to show that $\partial _xf(x,y)$ (resp. $\partial_yf(x,y)$ is continuous. This is true since $$\lim \limits_{(x,y)\to(0,0)} \partial_xf(x,y) = \lim \limits_{(x,y)\to(0,0)} \frac{4x^3}{2\sqrt{x^4+y^4}} = 0 = \partial_xf(0,0)$$ (analog for the partial to y). Since both partials exist and are continuous, we're done.
Now, is that attempt correct? If yes, did I do something I didn't necessarily have to do? Are there better/faster/easier methods to proof differentiability?
ADDITIONAL QUESTION: I did not prove that $\lim \limits_{(x,y)\to(0,0)} \frac{4x^3}{2\sqrt{x^4+y^4}} = 0$. Do I need to do that and how would I do that?

Answer 1

For the differentiability we have to consider $$L=\lim_{\left(x,y\right)\rightarrow\left(0,0\right)}\frac{\sqrt{x^{4}+y^{4}}}{\sqrt{x^{2}+y^{2}}} $$ and taking the polar coordinates we get

$$\color{red}{L=\lim_{\rho\rightarrow0^{+}}\frac{\rho^{2}\sqrt{\cos^{4}\left(\theta\right)+\sin^{4}\left(\theta\right)}}{\rho}=0.}$$

Answer 2

You could also prove $Df(0,0)=0$ directly. We need show

$$\frac{f(x,y) -[ f(0,0) + 0\cdot x + 0\cdot x]}{(x^2+y^2)^{1/2}} \to 0$$

as $(x,y) \to (0,0).$ Everything in brackets in the numerator is $0.$ So this fraction equals

$$\frac{(x^4+y^4)^{1/2}}{(x^2+y^2)^{1/2}} = \left (\frac{x^4+y^4}{x^2+y^2}\right )^{1/2}.$$

Since $(x^4+y^4)/(x^2 + y^2) \to 0,$ we're done.