Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Review - Exercises - Page 267: 9

Answer

$y'=\frac{ln(x)+1}{xln(x)}$

Work Step by Step

Start with the function: $y=ln(xln(x))$. Let $u = xln(x)$. Use the product rule to find u': $u'=1*ln(x)+x*\frac{1}{x} = ln(x)+1$. Substitute u into the equation: $y=ln(u)$. Use chain rule to differentiate: $y=\frac{u'}{u}$. Substitute expressions for u and u' back into the equation: $y'=\frac{ln(x)+1}{xln(x)}$.
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