Answer
g'(x)=\frac{-1}{x^{2}}
Work Step by Step
Prove inverse by showing f(g(x))=g(f(x))=x
f(g(x))=\frac{1}{1+\frac{1-x}{x}}=1\frac{1}{x}=x
g(f(x))=\frac{1-(\frac{1}{1+x})}{\frac{1}{1+x}}=\frac{x}{1+x}\times\frac{x+1}{1}=x
Compute g'(x) directly:
g'(x)=\frac{(1-x)'x-(1-x)x'}{x^{2}}=\frac{-x-1+x}{x^{2}}=\frac{-1}{x^{2}}
f'(x)=(x+1)^{-1}=-(x+1)^{-2}
f'(g(x))=-((\frac{1-x}{x})+1)^{-2}=-(\frac{1-x+x}{x})^{-2}=-(\frac{1}{x})^{-2}=-x^{2}
Therefore \frac{1}{f'(g(x))}=\frac{-1}{x^{2}}=g'(x)
