show that g is differentiable at x_0 and that f'(x_0)=g'(x_0)=h'(x_0)...?

Suppose that f,g,h are real valued functions on (a,b) and $\displaystyle x_0$$\displaystyle \in$(a,b). Assume that f and h are differentiable at $\displaystyle x_0$, f($\displaystyle x_0$)=h($\displaystyle x_0$) and f(x)<=g(x)<=h(x) for all x$\displaystyle \in$(a,b). show that g is differentiable at $\displaystyle x_0$ and f'($\displaystyle x_0$)=g'($\displaystyle x_0$)=h'($\displaystyle x_0$). Use this result to explain why the fraction F(x) defined by F(x)=$\displaystyle x^2$sin(1/x), x not equal to 0, and F(0)=0 has derivative equal to 0 at x=0.

I honestly have no clue :( I hope some one can help me thanks!