Calculus 8th Edition

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

Chapter 2 - Derivatives - 2.3 Differentiation Formulas - 2.3 Exercises: 34

Answer

$y'=\frac{-u^2+2u+8}{(1-u)^2}$

Work Step by Step

$y=\frac{(u+2)^2}{1-u}$ Use the quotient rule. Use the chain rule to derive the numerator. $y'=\frac{(1-u)2(u+2)(1)-(u+2)^2(0-1)}{(1-u)^2}$ Expand. $y'=\frac{(1-u)(2u+4)-(u^2+4u+4)(-1)}{(1-u)^2}$ Distribute. $y'=\frac{2u+4-2u^2-4u+u^2+4u+4}{(1-u)^2}$ Combine like terms. $y'=\frac{-u^2+2u+8}{(1-u)^2}$
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