Existence of directional derivative

[songoku]

Homework Statement

Please see below

Relevant Equations

Partial derivative

Direction derivative in the direction of unit vector u = <a, b>:
Du f(x,y) = fx (x,y) a + fy (x,y) b

My attempt:
I have proved (i), it is continuous since $\lim_{(x,y)\rightarrow (0,0)}=f(0,0)$

I also have shown the partial derivative exists for (ii), where $f_x=0$ and $f_y=0$

I have a problem with the directional derivative. Taking u = <a, b> , I got:
$$Du =\frac{\sqrt[3] y}{3 \sqrt[3] {x^2}}a+\frac{\sqrt[3] x}{3 \sqrt[3] {y^2}}b$$

Then how to check whether the directional derivative exists or not?

Thanks

 

[fresh_42]

The directional derivative at $[0,0]$ in direction $[a,b]$ (has to be zero since both partial directions are and) is given by
$$
D_{[0,0]}(f)\cdot [a,b]=0=\lim_{h\to 0}\dfrac{f([0,0]+h[a,b])-f([0,0])}{h}
$$
What do you get on the right-hand side for $f([x,y])= \sqrt[3]{xy}$?

 

[songoku]

The directional derivative at $[0,0]$ in direction $[a,b]$ (has to be zero since both partial directions are and) is given by
$$
D_{[0,0]}(f)\cdot [a,b]=0=\lim_{h\to 0}\dfrac{f([0,0]+h[a,b])-f([0,0])}{h}
$$
What do you get on the right-hand side for $f([x,y])= \sqrt[3]{xy}$?

$$0=\lim_{h\to 0}\frac{\sqrt[3]{(h^{2}ab)}-0}{h}$$
$$0=\lim_{h\to 0}\frac{(ab)^{\frac{1}{3}}}{h^{\frac{1}{3}}}$$

The limit can only exist if $(ab)^{\frac{1}{3}}=0$ so either $a=0$ or $b=0$ and since $u$ is unit vector, if $a=0$ then $b=\pm 1$ and vice versa

Is my working correct? Thanks

 

[fresh_42]

Yes, that is correct.

 

[songoku]

Thank you very much fresh_42