Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.6 - Derivatives of Logarithmic Functions - 3.6 Exercises - Page 223: 23


$y''=-\frac{1}{2}\frac{lnx}{2x\sqrt x}$

Work Step by Step

$y'=\frac{2+lnx}{2\sqrt x}$ and $y''=-\frac{1}{2}\frac{lnx}{2x\sqrt x}$ First derivative can be found by using product rule of differentiation. Here, $y'=\frac{d}{dx}(\sqrt xlnx)$ $=\sqrt x\frac{d}{dx}(lnx)+lnx\frac{d}{dx}{(\sqrt x)}$ $=\sqrt x.\frac{1}{x}+lnx.[\frac{1}{2}(x^{-1/2})]$ Thus, $y'=\frac{2+lnx}{2\sqrt x}$ After solving first derivative such as $y'=\frac{2+lnx}{2\sqrt x}$, we will find second derivative with the help of quotient rule of differentiation. $y''=\frac{1}{2}\frac{d}{dx}[\frac{2+lnx}{2\sqrt x}]$ $=\frac{\sqrt x\frac{d}{dx}(2+lnx)-(2+lnx)\frac{d}{dx}\sqrt x}{(\sqrt x)^{2}}$ $=-\frac{1}{2}\frac{lnx}{2x\sqrt x}$ Hence, $y''=-\frac{1}{2}\frac{lnx}{2x\sqrt x}$
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