Answer
$$g'(x)= (\frac{1}{1-x})^2$$
Work Step by Step
Since $f(x)=\frac{x}{x+1}$, then $f'(x)=\frac{1}{(x+1)^2}$. Hence, by Theorem 2 and assuming that $g(x)=f^{-1}(x)$, we have:
$y=\frac{x} {x+1}$
Switch $x$ and $y$:
$x=\frac{y}{y+1}$
Solve for $y$:
$x(y+1)=y$
$xy+x=y$
$x=y-xy$
$y=\frac{1}{1-x}$
Therefore, $$g(x)= \frac{x}{1-x}$$
We find $g'(x)$:
$$g'(x)=\frac{1}{f'(g(x))}=\frac{1}{1/(g(x)+1)^2}\\=\frac{1}{ ( 1/(1+x/(1-x))^2)}=(\frac{1}{1-x})^2$$
