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: 53

Answer

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

Work Step by Step

$f(x)=\frac{\ln(\frac{x^2}{x-1})}{\ln 2}$ $=\frac{1}{\ln 2}(2\ln x - ln (x-1))$ $f’(x)=\frac{1}{\ln 2} (\frac{2}{x}-\frac{1}{x-1})$ $=\frac{1}{\ln 2}\left(\frac{2x-2-x}{x(x-1)}\right)$ $=\frac{1}{\ln 2}\left(\frac{x-2}{x(x-1)}\right)$
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