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: 25


$g'(x)= \frac{1}{7}.$

Work Step by Step

Since $f(x)=7x+6$, then $f'(x)=7$. Hence, by Theorem 2, and assuming that $g(x)=f^{-1}(x)$, we have $$g'(x)=\frac{1}{f'(x)}=\frac{1}{7}.$$
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