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: 54



Work Step by Step

$$\frac{dy}{dt}=\frac{d}{dt}\Big(1+\cot(t/2)\Big)^{-2}$$ $$\frac{dy}{dt}=-2\Big(1+\cot(t/2)\Big)^{-3}\frac{d}{dt}(1+\cot(t/2))$$ $$\frac{dy}{dt}=-2\Big(1+\cot(t/2)\Big)^{-3}\Big(0-\csc^2(t/2)\frac{d}{dt}(t/2)\Big)$$ $$\frac{dy}{dt}=-2\Big(1+\cot(t/2)\Big)^{-3}\Big(-\frac{1}{2}\csc^2(t/2)\Big)$$ $$\frac{dy}{dt}=\csc^2(t/2)\Big(1+\cot(t/2)\Big)^{-3}$$
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