I have a question about a few manipulations when computing a directional derivative at the point $(0,0)$.
We have $f'(0,u) = \lim\limits_{h \to 0} \frac{f(hu)}{h}$ where $f(x,y) = (xy)^\frac{1}{3}$ and $u = (u_1, u_2)$. This is just the definition of a directional derivative.
Then we expand this to: $f'(0,u) = \lim\limits_{h \to 0} \frac{(h u_1h u_2)^\frac{1}{3}}{h}$
I don't understand the steps of this fully. Are the precise steps:
$f'(0,u) = \lim\limits_{h \to 0} \frac{f(hu)}{h} = \lim\limits_{h \to 0} \frac{f(h(u_1,u_2))}{h}=\lim\limits_{h \to 0} \frac{f((hu_1,hu_2))}{h}$
This seems to make sense to me but I just want to make sure I am allowed to do these operations inside the function (in particular, taking the scalar $h$ inside in the last step).
Thanks for your help.
Yes we have
$$f'(0,u) = \lim\limits_{h \to 0} \frac{f(hu)}{h} = \lim\limits_{h \to 0} \frac{f(h(u_1,u_2))}{h}=\lim\limits_{h \to 0} \frac{f((hu_1,hu_2))}{h}=\lim\limits_{h \to 0} \frac{h^\frac23(u_1u_2)^\frac13}{h}$$
and for $u_1,u_2 \neq 0$ the directional derivative doesn't exist.