The questions is as title and the function is defined as follows:

$f(x,y)= \frac{x^3y}{x^4+y^2}$ if $(x,y) \neq (0,0)$ and $0$ if $(x,y) = (0,0)$

Note: $u=(u_1, u_2)$

I tried to find the directional derivative by definition but I end up at $\frac {u_1^3}{u_2}$. Which is only showing that the directional derivative exists but clearly it is not equal to $0$.

And I can't seek to "those" theorems that allows me to calculate the directional derivatives using the dot product of the vector $u$ and the gradient of $f$ at $(0,0)$ because I know from my next part of the question it asks me to prove that $f$ is not differentiable at $(0,0)$.

Any help is appreciated, thanks in advance.


By definition,

$$\partial_uf(0,0)=\lim_{h\to 0} \frac{f(hu)-f(0,0)}{h}$$

We have $$\frac{f(hu)-f(0,0)}{h}=\frac{f(hu)}{h}=\frac{(hu_1)^3(hu_2)}{h[(hu_1)^4+(hu_2)^2]}=\frac{hu_1^3u_2}{h^2u_1^4+u_2^2}$$

If $u_2=0$, then $u_1=1$ and the expression above is zero for all $h\neq 0$, so the limit as $h\to0$ is zero. Otherwise we can just plug in $h=0$ to find the limit:

$$\partial_uf(0,0)=\lim_{h\to 0} \frac{hu_1^3u_2}{h^2u_1^4+u_2^2}=\frac{0}{0+u_2^2}=0$$

  • 1
    $\begingroup$ You forgot an $h$ on the denominator. $\endgroup$ – Skypanties Oct 23 '18 at 20:00
  • 1
    $\begingroup$ Thanks. I fixed it. $\endgroup$ – smcc Oct 23 '18 at 20:02
  • $\begingroup$ It seemed like I myself forgot an $h$ on the numerator... I feel stupid. Thanks though! At least you helped me catch my mistake. $\endgroup$ – Skypanties Oct 23 '18 at 20:03

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