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: 34

Answer

$U'(y)=\dfrac{10y(y^{4}+1)^{4}(y^{4}+2y^{2}-1)}{(y^{2}+1)^{6}}$

Work Step by Step

$U(y)=(\dfrac{y^{4}+1}{y^{2}+1})^{5}$ Differentiate using the chain rule: $U'(y)=5(\dfrac{y^{4}+1}{y^{2}+1})^{4}(\dfrac{y^{4}+1}{y^{2}+1})'=...$ Now, use the quotient rule to find $(\dfrac{y^{4}+1}{y^{2}+1})'$: $...=5(\dfrac{y^{4}+1}{y^{2}+1})^{4}[\dfrac{(y^{2}+1)(y^{4}+1)'-(y^{4}+1)(y^{2}+1)'}{(y^{2}+1)^{2}}]=...$ $...=5(\dfrac{y^{4}+1}{y^{2}+1})^{4}[\dfrac{(y^{2}+1)(4y^{3})-(y^{4}+1)(2y)}{(y^{2}+1)^{2}}]=...$ Simplify: $...=5(\dfrac{y^{4}+1}{y^{2}+1})^{4}[\dfrac{4y^{5}+4y^{3}-2y^{5}-2y}{(y^{2}+1)^{2}}]=5(\dfrac{y^{4}+1}{y^{2}+1})^{4}\dfrac{2y^{5}+4y^{3}-2y}{(y^{2}+1)^{2}}=\dfrac{10y(y^{4}+1)^{4}(y^{4}+2y^{2}-1)}{(y^{2}+1)^{6}}$
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