# Study the Frechet differentiability of a function.

Study the Frechet differentiability of this function at $(0,0)$

$$f(x,y)=\frac{x^3y}{(x^2+y^2)^{3/2}}$$ if $(x,y) \neq (0,0)$ and if it does then $f(x,y)=0$.

Attempted Solution:

I've found the vector of partial derivatives for this function and it is

$$<\frac{\partial f}{\partial x}=\frac{3x^2y^3}{(x^2+y^2)^{5/2}},\frac{\partial f}{\partial y}=\frac{x^5-2x^3y^2}{(x^2+y^2)^{5/2}}>$$

How does help me find my Frechet derivative? If it helps at all. My guess would be that the frechet derivative would be $F(x,y)=<\frac{3x^2y^3}{(x^2+y^2)^{5/2}},\frac{x^5-2x^3y^2}{(x^2+y^2)^{5/2}}>$ which is not defined at $(0,0)$ so the function is not Frechet differentiable. But this would just be a guess.

Any help is appreciated.

First note that since $f(x,0)\equiv 0$ and $f(0,y)\equiv 0$, the function $f$ has partial derivatives at $(0,0)$, both equal to $0$.
So, the only possible candidate for being the Fréchet derivative of $f$ at $(0,0)$ is the linear functional $L=0$.
Therefore, to determine whether $f$ is Fréchet-differentiable at $(0,0)$, the only thing you have to do is to check whether $f(x,y)=o(\Vert (x,y)\Vert)$ as $(x,y)\to (0,0)$. In particular, you should have $f(x,x)=o(\vert x\vert)$ as $x\to 0$; but you will check that this is not so.
• Thanks for the answer. What do you mean by $o(||(x,y)||)$ in particular what does $o$ mean? Commented Sep 26, 2013 at 17:41
• "$o({\rm something})$" means "negligible with respect to something". So $f(x,y)=o(\Vert (x,y)\Vert)$ means that $\frac{f(x,y)}{\Vert (x,y)\Vert}\to 0$ (as $(x,y)\to (0,0)$). Commented Sep 26, 2013 at 18:24