Answer
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 $f(x)$.
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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