Use uniform convergence to control the derivative of function Let $f_n$: $I:=[-1,1]\to \mathbb R$ be given, and we also assume that $f_n\in C^\infty(I)$.
Assume $f_n\to 0$ uniformly in $I$, and we know
$$
\limsup_{n\to\infty}\int_I\sqrt{1+(f'_n)^2}<\infty
$$
i.e., the length of the graph of $f_n$ is bounded.
My question: can we deduce that 
$$
\limsup_{n\to \infty}\,\left(\sup_{x\in I'}|f_n'(x)|\right)<\infty
$$
for some subinterval $I'\subset I$? How large $I'$ could be? i.e., can we prove that $\mathcal L^1(I\setminus I')=0$?
How about $f_n''$?

No, the derivatives need not bounded on any subinterval.  Consider 
$$
f_n(x) = \frac{1}{\sqrt{n}} (\cos  \sqrt{n}x  )^n
$$
The length of the graph is uniformly bounded, since there are about $\sqrt{n}$ waves of height $1/\sqrt{n}$. 
On the other hand, $$\sup \left| \frac{d}{dx} (\cos x)^n \right|\to \infty$$ (this is clear from the shape of the graph: when $n$ is large, there is a narrow peak of height $1$). The scaling by $\sqrt{n}$ does not change the supremum of derivative, but makes it attained more often. As a result, $f_n'$ becomes increasingly large increasingly often.
