group of diffeomorphisms of interval is perfect Every element in $\mathrm{Diff}([0,1])$, group of diffeomorphisms of interval fixing the endpoints, can be written as a product of commutators since this group is perfect (I don't know the proof though).
Is it possible to write every element as just one commutator?
Edit: @Jim Belk I'm confused actually. The motivation for this question was this statement ,which I've seen in a paper in a peer reviewed journal(without proof though): If S is a surface with boundary which is not compact planar, and $b$ is a boundary component and $\mu$ is a diffeomorphism of the interval, then there is a foliation of $S \times [0,1]$ (we can think about it as a fiber bundle over S with fiber [0,1]), transverse to [0,1] factor, such that it has holonomy $\mu$ on boundary $b$. I've considered the case that S is a once punctured torus because it implies the general case. Now if we look at representation of fundamental group into group of diffeomorphisms of the fiber ([0,1] here) then this implies that $\mu$ can be written as a commutator in the diffeomorphism group and these two statements are equivalent. Am I missing something?

It is not true that this group is perfect.  In particular, the function $D\colon \mathrm{Diff}([0,1])\to \mathbb{R}\times\mathbb{R}$ defined by
$$
D(f) \;=\; \bigl(\log f'(0),\log f'(1)\bigr)
$$
is a nontrivial homomorphism from $\mathrm{Diff}([0,1])$ onto an abelian group.
It turns out that the commutator subgroup of $\mathrm{Diff}([0,1])$ is precisely the kernel of this homomorphism.  See the following paper:
Fukui, Kazuhiko. “Homologies of the group $\mathrm{Diff}^\infty (\textbf{R}^n, 0)$ and its subgroups”. Journal of Mathematics of Kyoto University 20, no. 3 (1980): 475–487.
