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The answer by Bazin (https://mathoverflow.net/users/21907/bazin), Faa di Bruno's formula for vector valued functions, URL (version: 2012-09-04): https://mathoverflow.net/q/106339 is providing a formula for the $n$th derivative of $f\circ g$. Assuming that $g=\Phi$ is invertible and smooth, we take $f=\Phi^{-1}$ its inverse function and we find for $n\ge 2$ $$ 0=\sum_{\substack{n_1+\dots+n_r=n\\r\ge 1, n_j\ge 1}} \frac{f^{(r)}\circ g}{r!}\frac{g^{(n_1)}}{n_1!}\dots \frac{g^{(n_r)}}{n_r!}, $$ so that writing the "first" term for $r=n, n_j=1$, we get $$ \frac{f^{(n)}\circ g}{n!}\bigl(g^{(1)}\bigr)^n+\sum_{\substack{n_1+\dots+n_r=n\\n>r\ge 1, n_j\ge 1}} \frac{f^{(r)}\circ g}{r!}\frac{g^{(n_1)}}{n_1!}\dots \frac{g^{(n_r)}}{n_r!}=0, $$ and thus an explicit formula expressing $$ \bigl[(\Phi^{-1})^{(n)}\circ \Phi\bigr]\times \frac{(\Phi^{(1)})^n}{n!} $$ as a linear combination of the $r$th derivative of $\Phi^{-1}$ (with $1\le r<n$) whose coefficients are products $\frac{\Phi^{(n_1)}}{n_1!}\dots \frac{\Phi^{(n_r)}}{n_r!}$ where $n_1+\dots+n_r=n$. We get that $$ \underbrace{[(\Phi^{-1})^{(n)}\circ \Phi]\times \frac{(\Phi^{(1)})^n}{n!}}_{\Psi_n} =-\sum_{\substack{n_1+\dots+n_r=n\\n>r\ge 1, n_j\ge 1}} \underbrace{{[(\Phi^{-1})^{(r)}\circ \Phi]} \times \frac{(\Phi^{(1)})^r}{r!}}_{=\Psi_r} {(\Phi^{(1)})^{-r}} \frac{\Phi^{(n_1)}}{n_1!}\dots \frac{\Phi^{(n_r)}}{n_r!} , $$ that is $$ \Psi_n=-\sum_{n>r\ge 1} \Psi_r \underbrace{ {(\Phi^{(1)})^{-r}} \sum_{\substack{n_1+\dots+n_r=n\\ n_j\ge 1}} \frac{\Phi^{(n_1)}}{n_1!}\dots \frac{\Phi^{(n_r)}}{n_r!}}_{\Omega_{r,n}(\Phi)},\quad \text{and we have also } \Psi_1=I, $$ so that $$ \Psi_n=-\sum_{1\le r\le n-1}\Psi_r \Omega_{r,n}(\Phi), \quad \Psi_1= I.$$ We can now prove inductively that $\Vert\Psi_n\Vert$ is bounded above by a polynomial $$ P_n(\frac{\Vert \Phi^{(\nu_1)}\Vert}{\Vert \Phi^{(1)}\Vert }, \dots, \frac{\Vert \Phi^{(\nu_l)}\Vert}{\Vert \Phi^{(1)}\Vert }), \quad \nu_j\le n. $$