Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.9 Derivatives of Logarithmic and Exponential Functions - 3.9 Exercises - Page 211: 21


$\frac{d}{dx}\left(\frac{\ln(x)}{\ln(x)+1}\right) = \frac{1}{x(\ln(x)+1)^2}$

Work Step by Step

Using Chain Rule and Quotient Rule $\frac{d}{dx}(\frac{ln(x)}{ln(x)+1}) = \frac{(\frac{1}{x})(ln(x)+1)-(ln(x))(\frac{1}{x})}{(ln(x)+1)^2} = \frac{1}{x(\ln(x)+1)^2}$
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