## Calculus (3rd Edition)

$$g'(x)= (\frac{1}{1-x})^2$$
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 $g'(x)$: $$g'(x)=\frac{1}{f'(g(x))}=\frac{1}{1/(g(x)+1)^2}\\=\frac{1}{ ( 1/(1+x/(1-x))^2)}=(\frac{1}{1-x})^2$$