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I have the following function:
$f(x,y)=\frac{x^{2}y}{\sqrt{x^{2}+y^{2}}}$, when $(x,y)\neq (0,0)$ and

$f(x,y) =(0,0)$, when $(x,y)=(0,0)$.

I have to prove that f is totally differentiable, I tried doing this using the the theorem that $f$ is totally differentiable in the point $\xi $ if there exists a linear image $A$ such that:
$lim \frac{\| f(x)-f(\xi)-A(x-\xi)\|}{\|x-\xi\|}=0$, when $x\rightarrow \xi$.

I now substitute $A$ by the jacobi matrix of $f(\xi _{1},\xi _{2})$ to show that indeed this is true where the jacobi matrix is equal to:
$A=(\frac{\partial}{\partial \xi _{1}}f(\xi _{1},\xi _{2}),\frac{\partial}{\partial \xi _{2}}f(\xi _{1},\xi _{2}))=(\frac{\xi _{2}(\xi _{1}^{3}+2\xi _{1}\xi _{2}^{2})}{(\xi _{1}^{2}+\xi _{2}^{2})^{3/2}},\frac{\xi _{1}^{4}}{(\xi _{1}^{2}+\xi _{2}^{2})^{3/2}})$
and thus it schould hold that:
$lim \frac{\| f(x,y)-f(\xi_{1},\xi_{2})-A(\xi_{1},\xi_{2}).((x,y)-(\xi_{1},\xi_{2}))^{T}\|}{\|(x,y)-(\xi_{1},\xi_{2})\|}=0$, when $(x,y)\rightarrow (\xi_{1},\xi_{2})$
where we have that $(\xi_{1},\xi_{2})=(0,0)$, my problem is that $A(0,0)$ doesn't exist. I don't really know where to go from here, am I doing something wrong?

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1 Answer 1

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For this function there is an easier way to prove that it is totally differentiable. The only problem is at $(0,0)$.At any other point the partial derivatives are continuous so the function is differentiable. So at $(0,0)$: You can evaluate $\frac{\partial f}{\partial x}(0,0)=\lim_{x \to 0}\frac{f(x,0)-f(0,0)}{x}=0$ and $\frac{\partial f}{\partial y}(0,0)=0$. Then you evaluate the limit $\lim_{(x,y) \to (0,0)}\frac{f(x,y)-f(0,0)-\frac{\partial f}{\partial x}(0,0)x-\frac{\partial f}{\partial y}(0,0)y}{\sqrt {x^2+y^2}}=0$ .Therefore f is differentiable at $(0,0)$ and $df(0,0)(x_1,x_2)=0x_1+0x_2$

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