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Let $f:\mathbb{R}^2\rightarrow\mathbb{R}, (x,y)\mapsto\begin{cases}\frac{xy^2}{x^2+y^6},\,&(x,y)\neq(0,0)\\0,\,&\text{else}\end{cases}$. Show that for every direction $v\in\mathbb{R}^2\setminus\{0\}$ there exists a directional derivative $D_vf(0,0)$. Is $f$ continuous in $(0,0)$? Is $f$ differentiable in $(0,0)$?


I'm stuck on this exercise. I tried applying the definition, i.e. showing that $\lim_{t\rightarrow 0}\frac{f(\xi+tv)-f(\xi)}{t}$ exists, but it's getting me nowhere. Can anybody tell me the right approach here or what I'm not seeing?

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For the first part let consider for $\vec v=(a,b)$

$$\lim_{t\to 0} \frac{f(0+at,0+bt)-f(0,0)}{t}=\lim_{t\to 0} \frac{t^3ab^2}{t^3a^2+t^7b^6}=\lim_{t\to 0} \frac{ab^2}{a^2+t^4b^6}=\begin{cases}0\quad a=0\lor b=0\\\frac{b^2}a\quad a\neq 0\end{cases}$$

For the continuity let consider

$$\lim_{(x,y)\to (0,0)} \frac{xy^2}{x^2+y^6}$$

and observe that

  • $x=0 \implies \frac{xy^2}{x^2+y^6}=0$
  • $x=t^4 \quad y=t \quad t\to 0 \implies \frac{xy^2}{x^2+y^6}=\frac{t^6}{t^8+t^6}=\frac{1}{t^2+1}\to 1$

Finally recall that differentiability $\implies$ continuity.

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  • $\begingroup$ in your final observation, more information still, by using $x=t^3, y=t.$ I'm not always sure on these things so I use Lagrange multipliers to look for the maximum of the numerator constrained with the denominator some (small) constant. In this case it gave $6 y^6 = 4 x^2 \; ;$ then i threw out the constants. $\endgroup$ – Will Jagy Jul 1 '18 at 23:53
  • $\begingroup$ From the first part you already know that $f$ is not differentiable at the origin: the directional derivatives in two different directions are zero, so they all must be zero for there to be a consistent tangent plane. $\endgroup$ – amd Jul 2 '18 at 2:33
  • $\begingroup$ @amd I think that in this case you are not correct indeed differentiability implies existence of partial derivatives. Others implications are false. $\endgroup$ – user Jul 2 '18 at 5:44
  • $\begingroup$ If $f$ is differentiable and $f_x=f_y=0$ as you’ve shown, then all directional derivatives must vanish. $\endgroup$ – amd Jul 2 '18 at 7:30
  • $\begingroup$ @amd Ah ok now what you are claiming is clear to me! Yes of course we can also conlcude in that way. $\endgroup$ – user Jul 2 '18 at 7:39

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