Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.6 - The Chain Rule - Exercises 3.6 - Page 149: 54

Answer

$\frac{csc^2(\frac{t}{2})}{(1+cot(\frac{t}{2}))^3}$

Work Step by Step

y=$(1+cot(\frac{t}{2}))^{-2}$ Apply the chain rule: $\frac{dy}{dt}=-2(1+cot(\frac{t}{2}))^3.\frac{d}{dt}(1+cot(\frac{t}{2}))$ =$-2(1+cot(\frac{t}{2}))^{-3}.(-csc^2(\frac{t}{2})).\frac{d}{dt}(\frac{t}{2})$ =$\frac{csc^2(\frac{t}{2})}{(1+cot(\frac{t}{2}))^3}$
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