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: 54

Answer

$y’=\frac{1}{\ln 10}\left(\frac{x^2+1}{x(x^2-1)}\right)$

Work Step by Step

$y=\frac{\ln (\frac{x^2-3}{x})}{\ln 10}$ $=\frac{\ln (x^2-1)-\ln x}{\ln 10}$ $y’=\frac{1}{\ln 10}\left(\frac{2x}{x^2-1}-\frac{1}{x}\right)$ $y’=\frac{1}{\ln 10}\left(\frac{x^2+1}{x(x^2-1)}\right)$
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