Answer
$\frac{1-x^2}{x(x^2+1)}$
Work Step by Step
$f(x)=\frac{x}{x^2+1}$
$=ln x - ln(x^2+1)$
$f'(x)=\frac{1}{x}-\frac{2x}{x^2+1}$
$=\frac{x^2+1-2x^2}{x(x^2+1)}$
$=\frac{1-x^2}{x(x^2+1)}$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
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 - 5.1 Exercises - Page 325: 51
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