Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.5 The Chain Rule - 2.5 Exercises: 35

Answer

$$y'=\frac{16\sin2x(1-\cos2x)^3}{(1+\cos2x)^5}$$

Work Step by Step

$$y'=\left(\left(\frac{1-\cos2x}{1+\cos2x}\right)^4\right)'= 4\left(\frac{1-\cos2x}{1+\cos2x}\right)^3\left(\frac{1-\cos2x}{1+\cos2x}\right)'= 4\left(\frac{1-\cos2x}{1+\cos2x}\right)^3\frac{(1-\cos2x)'(1+\cos2x)-(1-\cos2x)(1+\cos2x)'}{(1+\cos2x)^2}= 4\left(\frac{1-\cos2x}{1+\cos2x}\right)^3\frac{-(-\sin2x)(2x)'(1+\cos2x)-(1-\cos2x)(-\sin2x)(2x)'}{(1+\cos2x)^2}=4\left(\frac{1-\cos2x}{1+\cos2x}\right)^3\frac{\sin2x\cdot2(1+\cos2x)+\sin2x(1-\cos2x)\cdot2}{(1+\cos2x)^2}= 4\left(\frac{1-\cos2x}{1+\cos2x}\right)^3\frac{2\sin2x+2\sin2x\cos2x+2\sin2x-2\sin2x\cos2x}{(1+\cos2x)^2}= 4\left(\frac{1-\cos2x}{1+\cos2x}\right)^3\frac{4\sin2x}{(1+\cos2x)^2}= \frac{16\sin2x(1-\cos2x)^3}{(1+\cos2x)^5}$$
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