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: 52

Answer

$f’(x)=\frac{2}{3(2x+1)\ln 2}$

Work Step by Step

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