# Show that $f(x,y)$ is differentiable at $(0,0)$

Show that $f(x,y)$ defined by:

$$f(x,y) = \begin{cases}\dfrac{x^2y^2}{\sqrt{x^2+y^2}}&\text{ if }(x,y)\not =(0,0)\\0 &\text{ if }(x,y)=(0,0)\end{cases}$$

is differentiable at $(x,y) = (0,0)$

I tried to solve this problem by applying the theorem that if partial derivatives are continuous then the function is differentiable. Therefore, I calculated partial derivated but not I am stuck in showing they are indeed continuous. Help me!

• Hint: $\sqrt{x^2+y^2} \ge \max\{|x|, |y|\}$ and $|xy| =\min\{|x|, |y|\} \max\{|x|, |y|\}$ – user251257 Oct 18 '15 at 3:46
• The title and the problem do not match. Your title says that "show that $f$ is differentiable at $(0,0)$", however you haven't specified the value of $f(0,0)$. In fact, if $f(0,0) \ne 0$, then $f$ is not even continuous at $(0,0)$ and can not be differentiable at $(0,0)$. – Mercy King Oct 18 '15 at 4:12
• If $f(0,0)=0$, then $df(0)\equiv0$ because $|f(h)|\le \|h\|_2^3$ for all $h \in \mathbb{R}^2$ – Mercy King Oct 18 '15 at 4:20
• @MercyKing I edited the problem. Sorry for the confusion! – Guten Tag Oct 18 '15 at 4:48
• @user251257 I didn't get your hint. Can you explain furthur please? – Guten Tag Oct 18 '15 at 4:51

You have $$x^2+y^2-2 \vert xy \vert=(\vert x \vert - \vert y \vert)^2 \ge 0$$ Hence $$\vert xy \vert \le \frac{x^2+y^2}{2}$$ and $$0 \le \frac{\vert f(x,y) \vert}{\sqrt{x^2+y^2}} = \frac{x^2y^2}{x^2+y^2} \le \frac{1}{4}(x^2+y^2)$$ As $\lim_{(x,y) \to (0,0)} x^2+y^2 = 0$, this proves that $f$ is differentiable at $(0,0)$ and that its Fréchet derivative is equal to $0$. Which means that $f_x(0,0)=f_y(0,0)=0$.
For every non-zero $h=(h_1,h_2)\in \mathbb{R}^2$ we have: $$|f(h)-f(0)|=||f(h)|=\frac{h_1^2h_2^2}{\|h\|_2}\le \frac{\|h\|_2^4}{\|h\|_2}=\|h\|_2^3,$$ and therefore $$\lim_{\|h\|_2\to0}\frac{|f(h)-f(0)|}{\|h\|_2}=0.$$ This shows that $f$ is differentiable at $(0,0)$, and $Df(0)\equiv 0$.