Let $$f(x,y)= \begin{cases} \sin\left(\frac{y^2}{x}\right)\sqrt{x^2+y^2}, & x\neq0\\ 0,&x=0 \end{cases}$$

Then, is $f$ differentiable at the origin? I think no, but I also find that the directional derivatives exists at all points and is equal to 0, which is a linear function. Any hints. Thanks beforehand.

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    $\begingroup$ What is the definition of $f$ in points of the form $(0,t)$ ?? $\endgroup$ – Fred Jan 31 '18 at 7:46
  • $\begingroup$ @Fred edited the post. Please check now $\endgroup$ – vidyarthi Jan 31 '18 at 7:57

You're correct that all directional derivatives vanish at the origin. (But there are discontinuous functions that have that property!) Here's a hint to answer your question: Is $$\lim_{(x,y)\to (0,0)} \sin\big(\tfrac{y^2}x\big) = 0?$$

  • $\begingroup$ I suspected it. If we approach the origin via the parabola $y^2=x$, we obtain that the function is discontinuous. Am I right in that? But, the directional derivative is linear, right? Also, how to best approach such kind of problems? Any standard algorithm? $\endgroup$ – vidyarthi Jan 31 '18 at 8:05
  • $\begingroup$ No standard algorithm other than to find the partial derivatives and hence the only possible candidate for the derivative matrix ... and then check the limit definition. Of course, if you can see the function is discontinuous, then of course it cannot be differentiable. $\endgroup$ – Ted Shifrin Jan 31 '18 at 8:08

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