0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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:

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

$\endgroup$
  • 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

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.