Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.2 Inverse Functions - Exercises - Page 335: 29


$$g'(x)= (\frac{1}{1-x})^2$$

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.