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: 28


$$ g'(x)= \frac{1}{12}(\frac{4}{x+1})^{2/3} .$$

Work Step by Step

Since $f(x)=4x^3-1$, then $f'(x)=12x^2$. Hence by Theorem 2 and assuming that $g(x)=f^{-1}(x)$, we have $$g(x)= (\frac{x+1}{4})^{1/3}, \quad g'(x)=\frac{1}{f'(g(x))}=\frac{1}{12((x+1)/4))^{2/3} }=\frac{1}{12}(\frac{4}{x+1})^{2/3} .$$
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