3

Based of Wikipedia -> Requirement of continuity I'm trying to figure out why the second derivative of the function is not symmetric.

The function is:

$$ f(x,y) = \begin{cases} \frac{xy(x^2 - y^2)}{x^2+y^2} & \mbox{ for } (x, y) \ne (0, 0)\\ 0 & \mbox{ for } (x, y) = (0, 0). \end{cases} $$

(1) I want to be sure that i'm right doing the the first derivative at (0,0) $$ {\partial _xf(0,0) = } \lim_{t \to 0} \frac{f(0+t,0) - f(0,0)}{t} = 0 $$

(2) I don't understand how they obtain the second derivative, they say the second partial derivatives are not continuous at (0,0), and the symmetry fails.

Can you explain how to obtain the second derivatives $\partial _x \partial _yf$ and $\partial _y \partial _xf$ at (0,0) and why the symmetry fails.

Thank you in advance.

andres
  • 31
  • The first derivative is correctly computed. The second partial derivative at $(0,0)$ is just $\lim_{t \to 0} \frac{f(0,0+t) - f(0,0)}{t}$. This will also be zero. Try computing the partial derivatives at a general point. – Sarvesh Ravichandran Iyer Mar 23 '17 at 00:57
  • First of all, there's something wrong with your notation in formula (1): what is t? Did you mean $t \rightarrow 0$ instead of $h \rightarrow (0,0)$? – Jay Zha Mar 23 '17 at 00:58
  • Wikipedia say $ \partial _x \partial _yf = 1 $ and $ \partial _y \partial _xf = -1 $ at (0,0) – andres Mar 23 '17 at 01:07

0 Answers0