Let $f:\mathbb{R}^n\to\mathbb{R}$. Lets assume that for every differetiable curve $\gamma:[-1,1]\to\mathbb{R}^n$ where $\gamma(0)=\underline{0}$, $f\circ\gamma[-1,1]\to\mathbb{R}$ differentiable at $t=0$. Prove that $f$ is differentiable at $\underline{0}$.
Denote: $g = f\circ \gamma$. Let $\gamma$ as described. By the chain-rule: $$g'(0) = D_f(\gamma(0))\gamma'(0) = 0$$.
But, can I have even do that? I mean, we don't know that $f$ is differetiable yet and I just used $D_f$.
More thoughts:
- Choosing a specific $\gamma$.
- Looking at the function from both $[-1, 0]$ and $[0,1]$.
Heck, lots of thoughts but couldn't put my finger on the solution.