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) $
$g’(x) = \frac{-e^x}{(1+e^x)^2}$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 3 - Differentiation - 3.3 Product and Quotient Rules - Exercises - Page 121: 11
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