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 164: 49

Answer

$$ y'= \frac{8\csc^2\theta}{(1+\cot \theta)^2}.$$

Work Step by Step

Recall that $(\cot x)'=-\csc^2 x$. Since $ y=\frac{8}{1+\cot \theta }$, then by the quotient rule $(u/v)'=\frac{vu'-uv'}{v^2}$, the derivative $ y'$ is given by $$ y'= \frac{(1+\cot \theta) (0)-8(-\csc^2\theta)}{(1+\cot \theta)^2}\\ =\frac{8\csc^2\theta}{(1+\cot \theta)^2}.$$
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