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 - Page 158: 44


$$y'=-12\cos^3(\sin^3x)\sin(\sin^3x)\sin^2x\cos x$$

Work Step by Step

$$y'=(\cos^4(\sin^3x))'=4\cos^3(\sin^3x)\cdot(\cos(\sin^3x))'= 4\cos^3(\sin^3x)\cdot(-\sin(\sin^3x))\cdot(\sin^3x)'= -4\cos^3(\sin^3x)\cdot\sin(\sin^3x)\cdot3\sin^2x\cdot(\sin x)'= -12\cos^3(\sin^3x)\sin(\sin^3x)\sin^2x\cos x$$
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