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: 24

Answer

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

Work Step by Step

$$U'(y)=\left(\left(\frac{y^4+1}{y ^2+1}\right)^5\right)'=5\left(\frac{y^4+1}{y^2+1}\right)^4\left(\frac{y^4+1}{y^2+1}\right)'=5\left(\frac{y^4+1}{y^2+1}\right)^4\frac{(y^4+1)'(y^2+1)-(y^4+1)(y^2+1)'}{(y^2+1)^2}= 5\left(\frac{y^4+1}{y^2+1}\right)^4\frac{4y^3(y^2+1)-(y^4+1)\cdot2y}{(y^2+1)^2}=5\left(\frac{y^4+1}{y^2+1}\right)^4\frac{4y^5+4y^3-2y^5-2y}{(y^2+1)^2}=5\left(\frac{y^4+1}{y^2+1}\right)^4\frac{2y^5+4y^3-2y}{(y^2+1)^2}=\frac{10y(y^4+2y^2-1)(y^4+1)^4}{(y^2+1)^6}$$
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