# Thread: deferentiability question ts7 q1

1. ## deferentiability question ts7 q1

i have this function
$\displaystyle f(x,y)=\begin{cases}x^{2}+y^{2}ln(x^{2}+y^{2}) & (x,y)\neq(0,0)\end{cases}$
and f(x,y)=0 on point (0,0)

i dont know the proper english term
i need to prove that f(x,y) is deferentialbe

so by the definition
$\displaystyle f(x,y)-f(0,0)-f_{x}(x-0)-f_{y}(y-0)=\epsilon_{1}(x-0)+\epsilon_{2}(y-0)$
and i need to find those epsilons which will make the sude to be equal
and both epsilons will go to zeroo when x and y are going to zero

i dont know what to put instead of "f" in each term
because the function is stplitted

2. ## Re: deferentiability question ts7 q1

Originally Posted by transgalactic
i have this function
$\displaystyle f(x,y)=\begin{cases}x^{2}+y^{2}ln(x^{2}+y^{2}) & (x,y)\neq(0,0)\end{cases}$
and f(x,y)=0 on point (0,0)

i dont know the proper english term
i need to prove that f(x,y) is deferentialbe

so by the definition
$\displaystyle f(x,y)-f(0,0)-f_{x}(x-0)-f_{y}(y-0)=\epsilon_{1}(x-0)+\epsilon_{2}(y-0)$
and i need to find those epsilons which will make the sude to be equal
and both epsilons will go to zeroo when x and y are going to zero

i dont know what to put instead of "f" in each term
because the function is stplitted

What you did? I really don't understand...

First check if f(x,y) is continuous in point (0,0). If it isn't therefor you can conclude that f(x,y) is not deferentiable.

Edit:
Are you sure about the function, perhaps you meant to write: (x^2+y^2)ln(x^2+y^2) ?

3. ## Re: deferentiability question ts7 q1

this is a question of checking by definition

4. ## Re: deferentiability question ts7 q1

Originally Posted by transgalactic
this is a question of checking by definition

Perhaps the function is

$\displaystyle f(x,y)=\begin{Bmatrix} (x^2+y^2)\ln (x^2+y^2) & \mbox{ if }& (x,y)\neq (0,0)\\0 & \mbox{if}& (x,y)=0\end{matrix}$

Then, $\displaystyle \frac{\partial f}{\partial x}(0,0)=\frac{\partial f}{\partial y}(0,0)=\ldots=0$ so, if $\displaystyle f$ is differentiable at $\displaystyle (0,0)$ the only possible differential is $\displaystyle \lambda:\mathbb{R}^2\to \mathbb{R},\;\lambda (h)=0$ . Now you have to study the limit

$\displaystyle \lim_{h \to 0}{\frac{|f(h)-f(0)-\lambda(h)|}{ \left\|{h}\right\|}}=\lim_{h \to 0}{\frac{|f(h)|}{ \left\|{h}\right\|}}$

The function is differentiable at $\displaystyle (0,0)$ iff the limit is $\displaystyle 0$ .

5. ## Re: deferentiability question ts7 q1

i need to use only the epsilon method

anybody knows it?