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: 46

Answer

$$y'=4(x+(x+\sin^2x)^3)^3(1+3(x+\sin^2x)^2(1+\sin2x))$$

Work Step by Step

$$y'=\left((x+(x+\sin^2x)^3)^4\right)'= 4(x+(x+\sin^2x)^3)^3\cdot((x+(x+\sin^2x)^3)'= 4(x+(x+\sin^2x)^3)^3\cdot(1+3(x+\sin^2x)^2\cdot(x+\sin^2x)')= 4(x+(x+\sin^2x)^3)^3(1+3(x+\sin^2x)^2(1+2\sin x\cdot(\sin x)'))= 4(x+(x+\sin^2x)^3)^3(1+3(x+\sin^2x)^2(1+2\sin x\cos x))= 4(x+(x+\sin^2x)^3)^3(1+3(x+\sin^2x)^2(1+\sin2x))$$
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