-1

Consider the following function given by;

$f(x,y) = \begin{cases} \frac{xy}{x^2+y^2},&(x,y) \neq 0 \\\\ \ \ \ \ 0,&(x,y)=0 \end{cases}$

Is the function differentiable in the ordinate pair $(0,0)$???

Prove it

I calculated the following derivative.

$$ \frac{\lambda f}{\lambda y}(0,0)=\lim_{h\to 0}\frac{f(0+h,0)-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{(0+h)-0}{(0+h)^2+0^2}-0}{h}=\lim_{h\to 0}\frac{0}{h^3}=0$$

considering $(0,0)$

$$ \frac{\lambda f}{\lambda y}(0,0)=\lim_{h\to 0}\frac{f(0,0+h)-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{0(0+h)}{0^2+(0+h)^2}-0}{h}=\lim_{h\to 0}\frac{0}{h^3}=0$$

I could not differentiate the derivative

Replica of Show discontinuity of $\frac{xy}{x^2+y^2}$ I need another answer

Community
  • 1
Enrique García Presenda
  • 5

2 Answers2

0

Let $y = tx$ then $f(x,tx) = {t x^2 \over (1+t^2) x^2} = { t \over 1+t^2}$.

Choose $t = 0$ and let $x \to 0$, the limit is $0$.

Choose $t = 1$ and let $x \to 0$, the limit is ${1 \over 2}$.

copper.hat
  • 172,524
0

AT first of all it is not continuous. Consider the sequence of points $(\frac{1}{n},\frac{1}{n})$ the values of the functions are constant i.e. $\frac{1}{2}$ , but to be continuous its limit as $ n \rightarrow \infty $ it has to be zero . So it is not continuous . Hence not differentiable.

Chirantan Chowdhury
  • 1,088