Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 40


$$ y'=\frac{-\csc^2 x}{\cot x}.$$

Work Step by Step

Recall that $(\ln x)'=\dfrac{1}{x}$ Recall that $(\cot x)'=-\csc^2 x$. Since $ y=\ln \cot x $, then by the chain rule, we have $$ y'=\frac{-\csc^2 x}{\cot x}.$$
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