The problem is:
Let $f,g:\mathbb{R}\rightarrow \mathbb{R}$, whith $n$ derivates. If exist $a\in \mathbb{R}$ with $f'(a)=\cdots =f^{(n)}(a)=0$ then $(g\circ f)^{(i)}=0$ for $i=1,...,n$
I think in use inducction for find a general expresion of $(g\circ f)^{(i)}$ and show explicit that all members are multiplicated by any derivate of $f$ but is hard get a general expresion. I foun in a book the next exercise:
If $f,g$ are functions in the same conditions then for any naturals $n_1,...,n_k$ with $n=n_1+\cdots+n_k$ exist an integer number $\alpha=\alpha(n_1,...,n_k)$
$$(g\circ f)^{(n)}=\sum_{k=1}^n \alpha(n_1,...,n_k)\cdot (g^{(k)}\circ f)\cdot f^{(n_1)}\cdot f^{(n_2)}\cdots f^{(n_k)}$$
where for each $k=1,...,n$ is valid $n=n_1+\cdots+n_k$
This theorem helps to show the dependence of the drivates of $f$ but I can't do this exercise.
Thanks for help!