Answer
$f'(x)=\frac{e^x}{1+e^{2x}}$
Work Step by Step
$\arctan(\frac{e^x}{1})$
$let $
$\frac{e^x}=u, u'= e^{x}$
$\frac{d}{dx}(\arctan u)=\frac{u'}{1+u^2}$
$f'(x)=\frac{e^x}{1+e^{2x}}$
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 - 5.6 Exercises - Page 372: 43
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