I am now dealing with a problem which using Cauchy Mean Value Theorem, Let $f$ be a $2n+2$ continuously differentiable function over $[-n,n]$ and $p(x)$ a polynomial of degree $\leq 2n+1$ by interpolate $f$ at integer points, I also let $Q(x)=x(x^2-1^2)(x^2-2^2)\cdots (x^2-n^2)$. I have shown that there is $\xi_x\in(-n,n)$ such that $$L(x)=\dfrac{f(x)-p(x)}{Q(x)}=\dfrac{f^{(2n+1)}(\xi_x)}{(2n+1)!}.$$ For each fixed $x$, if I choose $\xi(x)=\min\{\xi_x\}$, then $$L(x)=\dfrac{f^{(2n+1)}(\xi(x))}{(2n+1)!}$$ Somehow I need to find the derivative $L(x)$ in terms of $f^{(2n+2)}$ and $\xi'$, but I don't know whether $\xi$ is differentiable or not. Therefore I simplify the question into:
Question: If $f\circ g$ and $f$ are differentiable function over $\mathbb R$, is $g$ differentiable over $\mathbb R$?
It is also possible that my way to prove the result is false, so if this is not always true, can you give a counterexample?
No, it is not. Take $f(x)=1$ for all $x$, and take $g(x)=1$ if $x$ is rational and $g(x)=-1$ if $x$ is irrational. $f$ and $f\circ g$ are both constant, but $g$ is nowhere continuous.