## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 3: Derivatives - Section 3.6 - The Chain Rule - Exercises 3.6 - Page 149: 53

#### Answer

$\frac{8 \sin 2t}{(1+cos2t)^5}$

#### Work Step by Step

y=$(1+cos2t)^{-4}$ Apply the chain rule: $\frac{dy}{dt}$$=-4(1+cos2t)^{-5}.\frac{d}{dt}(1+cos2t)$ =$-4(1+cos2t)^{-5}(-sin2t).\frac{d}{dt}(2t)$ =$\frac{8 \sin 2t}{(1+cos2t)^5}$

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