Calculus (3rd Edition)

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

Chapter 3 - Differentiation - Chapter Review Exercises - Page 163: 48


$$ y'= \csc(9z+1)-9z\csc (9z+1) \cot (9z+1).$$

Work Step by Step

Recall the product rule: $(uv)'=u'v+uv'$ Recall that $(\csc x)'=-\csc x \cot x$. Since $ y=z\csc(9z+1)$, then by the product rule, the derivative $ y'$ is given by $$ y'= \csc(9z+1)+9z(-\csc (9z+1) \cot (9z+1))\\= \csc(9z+1)-9z\csc (9z+1) \cot (9z+1).$$
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