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: 51


$$ y'= - \frac{-\csc \theta \cot \theta}{2\sqrt{1+\csc \theta }} \sec^2\sqrt{1+\csc \theta }.$$

Work Step by Step

Recall that $(\tan x)'=\sec^2 x$. Recall that $(\csc x)'=-\csc x \cot x$. Since $ y=\tan\sqrt{1+\csc \theta }$, then by the chain rule, the derivative $ y'$ is given by $$ y'= \sec^2(\sqrt{1+\csc \theta } ) \left(\frac{-\csc \theta \cot \theta}{2\sqrt{1+\csc \theta }}\right)\\=- \frac{-\csc \theta \cot \theta}{2\sqrt{1+\csc \theta }} \sec^2\sqrt{1+\csc \theta }.$$
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