Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.5 Exercises - Page 362: 55


$=\frac{1}{\ln 3}(\frac{3x-2}{2x(x-1)})$

Work Step by Step

$h(x)=\frac{\ln\left(\frac{x(x-1)^{\frac{1}{2}}}{2}\right)}{\ln 3}$ $=\frac{\ln x+ \frac{1}{2} \ln(x-1) - \ln 2}{\ln 3}$ $h’(x)= \frac{1}{\ln 3}(\frac{1}{x}+\frac{1}{2x-2}-0)$ $=\frac{1}{\ln 3}(\frac{3x-2}{2x(x-1)})$
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