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: 24


$$ h'(z)=(1+\frac{1}{z})\sec(z+\ln z)\tan(z+\ln z).$$

Work Step by Step

Recall that $(\ln x)'=\dfrac{1}{x}$ Recall that $(\sec x)'=\sec x\tan x$. Since $ h(z)=\sec(z+\ln z)$, then the derivative, using the chain rule, is given by $$ h'(z)=\sec(z+\ln z)\tan(z+\ln z)(z+\ln z)'=(1+\frac{1}{z})\sec(z+\ln z)\tan(z+\ln z).$$
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