University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 158: 48


$$q'=\csc^2\Big(\frac{\sin t}{t}\Big)\frac{\sin t-t\cos t}{t^2}$$

Work Step by Step

$$q=\cot\Big(\frac{\sin t}{t}\Big)$$ The derivative of function $q$ is $$q'=-\csc^2\Big(\frac{\sin t}{t}\Big)\Big(\frac{\sin t}{t}\Big)'=-\csc^2\Big(\frac{\sin t}{t}\Big)\frac{(\sin t)'t-\sin t(t)'}{t^2}$$ $$q'=-\csc^2\Big(\frac{\sin t}{t}\Big)\frac{t\cos t-\sin t}{t^2}$$ $$q'=\csc^2\Big(\frac{\sin t}{t}\Big)\frac{\sin t-t\cos t}{t^2}$$
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