Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 386: 12


See the proof below.

Work Step by Step

Assume that $f'(x)=f(x)^2$, then we have $$g'(x)=\frac{1}{f'(g(x))}=\frac{1}{f(g(x))^2}=\frac{1}{x^2}$$ where $g(x)$ is the inverse of $f(x)$.
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