# Show that for all unit vector $u$, the directional derivative $\partial_uf(0,0)$ exists and is zero.

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:

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

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