Answer
$\dfrac{1}{3}$
Work Step by Step
Here, $f'(x)= \dfrac{d}{dx}[e^x+x]=e^x+x$
Now, $| \dfrac{df^{-1}}{dx}|_{x=f(\ln 2)}=\dfrac{1}{f'[f^{-1}(f\ln 2))]}$
This implies that
$\dfrac{1}{e^{\ln2}+1}=\dfrac{1}{2+1}=\dfrac{1}{3}$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 7 - Practice Exercises - Page 420: 19
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