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: 44

Answer

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

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.