#### Answer

$g’(x) = \frac{-e^x}{(1+e^x)^2}$

#### Work Step by Step

$g(x) = \frac{1}{1+e^{x}}$
$g(x) = (1+e^{x})^{-1}$
$Power$ $Rule:$
$g’(x) = (-1)(1+e^x)^{-2}(e^x) $
$g’(x) = \frac{-e^x}{(1+e^x)^2}$

Published by
W. H. Freeman

ISBN 10:
1464125260

ISBN 13:
978-1-46412-526-3

$g’(x) = \frac{-e^x}{(1+e^x)^2}$

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