Show that the function defined by $$f(x,y) = \Bigg\{ \frac{xy(x^2-y^2) }{x^2+y^2}, \space when \space (x,y) \neq (0,0)$$ and $$f(x,y) = 0, \space when \space (x,y) =(0,0)$$

is differentiable at the origin

I am doing calculus course as part of graduation.

I followed the approach given in answer to following question.

Can the function $f(x,y) = \frac{xy}{\sqrt{x^2+y^2}}$ be defined so that f is differentiable at the origin?

I found $f_x$ and $f_y$. They exist and equal to zero.

I reached the following limit

$$\lim_{(x,y) \rightarrow 0} \frac{xy(x^2-y^2)}{(x^2+y^2)||(x,y)||}$$

I couldn't simplify this limit.

Please help me to simplify this limit.



Use polar coordinates: $$ \frac{xy(x^2-y^2)}{(x^2+y^2)\|(x,y)\|}=\frac{r^4\cos\theta\sin\theta(\cos^2\theta-\sin^2\theta)}{r^3}=\dotsm$$

  • $\begingroup$ Are you suggesting to use $|f(x,y)-f(0,0)| < \epsilon$ definition to find the limit $\endgroup$ – user493838 Oct 21 '17 at 18:11
  • $\begingroup$ I just answered your question about simplifying the limit. For differentiability, a sufficient condition is the continuity of partial derivatives. $\endgroup$ – Bernard Oct 21 '17 at 18:17
  • $\begingroup$ @Bernard You've shown the limit equals $0$ and that proves $Df(0,0)$ is the zero linear transformation. $\endgroup$ – zhw. Oct 22 '17 at 7:54

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