If $f: (-1,1)→M_n(\mathbb R)$ be a $C^1$ map such that $f(0)=I$ and $f(t)\in O(n)$ ,show that $Df(0)$ exists and is a skew-symmetric matrix

Question

If $f: (-1,1)→M_n(\mathbb R)$ be a $C^1$ map such that $f(0)=I$ and $f(t)\in O(n)$ ,show that $Df(0)$ exists is skew-symmetric . matrix

Actually, we have $f(t).f(t)^T=I$ for all $t,$ and we are to find the limit $\frac{\|f(h)-f(0)\|}{h}$ , but how to differentiate $f(t).f(t)^T=I$ in both sides ??

$f(t)$ can be thought as points in $\mathbb R^{n^2}$ and applying derivative component-wise, we have $f(t).f'(t)^T + f'(t).f(t)^T =0$ and by using $f(0)=I$ ,we get the desired result.

Now, is the converse true? For any skew-symmetric matrix $A$, does there exist a curve $g$ with $g(0)=I$ and $Dg(0)=A?$

I think $g(t)=exp(tA)$ will work

A small hint is required , thanks in advance.