Let $f:\mathbb{R}^m\rightarrow\mathbb{R}^n$ be Frechet differentiable. Show carefully that if $f'$ has directional derivative $D_u f'(a)$ for some $a\in \mathbb{R}^m$ and $u\neq 0$, then for every $v\neq 0$ ,the directional derivative $D_uD_v f(a)$ exists and equals $D_uf'(a) (v)$
Please do not post a solution to this problem, you only have to say why I'm right or wrong to get your points
The "show carefully" has thrown me because I think I have two just too easy to believe ways of proving the statement. I can't convince myself that they're right.
Way 1:
$$D_u f'(a) v=D_u[f'(a)v]=D_u[D_vf(a)]=D_uD_vf(a)$$
I doubt I can use associativity in this way, thats probably wrong. But then I have a slightly more roundabout way of doing the same thing,
Way 2:
$$D_u f'(a)=\lim_{t\rightarrow 0}\frac{f'(a+tu)-f'(a)}{t} \\ \iff t D_uf'(a)+o(t)=f'(a+tu)-f'(a)\\\ f'(a+tu)=f'(a)+tD_uf'(a)+o(t)$$ Apply the operator to $v$, $$ f'(a+tu)v=f'(a)v+t[D_uf'(a)]v+o(t)$$ Use $D_vf(x)=f'(x)v$ then $$D_v f(a+tu)=D_vf(a)+t D_uf'(a)v+o(t) $$ which is the defintion of the partial derivative so done?