Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 46

Answer

$$y'= 4{[x+{(x+\sin^2 x)}^3]}^3 [3{(x+\sin^2 x)}^2 (1+2\sin x\cos x)+1]$$

Work Step by Step

$y' = \frac{d}{dx} {[x+{(x+\sin^2 x)}^3]}^4$ By the chain rule: $= \frac{d{[x+{(x+\sin^2 x)}^3]}^4}{d[x+{(x+\sin^2 x)}^3]} \times \frac{d[x+{(x+\sin^2 x)}^3]}{dx}$ $= 4{[x+{(x+\sin^2 x)}^3]}^3 \times [\frac{d}{dx} [x]+\frac{d}{dx}[{(x+\sin^2 x)}^3]]$ Using the chain rule again: $= 4{[x+{(x+\sin^2 x)}^3]}^3 \times [1+\frac{d[{(x+\sin^2 x)}^3]}{d[{(x+\sin^2 x)}]} \times\frac{d[{(x+\sin^2 x)}]}{dx}]$ $= 4{[x+{(x+\sin^2 x)}^3]}^3 \times [1+3{(x+\sin^2 x)}^2 (1+2\sin x\cos x)]$ $= 4{[x+{(x+\sin^2 x)}^3]}^3 [3{(x+\sin^2 x)}^2 (1+2\sin x\cos x)+1]$
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