$f:\mathbb{R}^2\rightarrow \mathbb{R}$ Defined by $$f(x,y)= \frac{xy^2}{x^2+y^4}$$ if $x\neq 0,y\in\mathbb{R}$ and $$f(x,y)=0$$ if $x=0,y\in\mathbb{R}$


  1. it is continuous but not differentiable at origin

  2. differentiable at origin

  3. has all first order partial derivative at origin.

  4. does not have all first order derivatives at origin.

consider the limit $(x,y)\rightarrow (0,0)$ along the curve $y=m\sqrt{x}$ we get lim$$(x,y)\rightarrow(0,0)\frac{x^2m^2}{x^2+m^4x^2}=\frac{m^2}{1+m^4}$$ which is different for different values for $m$ hence $f$ is not continuous at origin, so 1 is false, and 2 is clearly false. I have checked that $f_x$ and $f_y$ exists at $(0,0)$ so only $3$ is correct and all others are false. could any one confirm me am I right? Thank you.

  • $\begingroup$ The question "What more can I say" has infinitely many distinct answers, I suppose :-) Can you make it a bit more precise? $\endgroup$ – Siminore Jul 12 '12 at 11:47
  • $\begingroup$ @Siminore edited :) $\endgroup$ – Marso Jul 12 '12 at 11:54
  • 2
    $\begingroup$ I think the titled should be more telling. $\endgroup$ – Quinn Culver Jul 12 '12 at 11:57
  • $\begingroup$ By $f(x)$ do you mean $f(x,y)$? $\endgroup$ – Mercy King Jul 12 '12 at 11:59
  • $\begingroup$ edited @Mercy thank you $\endgroup$ – Marso Jul 12 '12 at 12:05

$\displaystyle \lim_{x\rightarrow0}\frac{f(x,0)-f(0,0)}x$is found to be 0. Similarly for $y$. So 3 is true and not 4

  • $\begingroup$ thank you aneesh got it. :) $\endgroup$ – Marso Jul 12 '12 at 11:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.