Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.6 - Derivatives of Logarithmic Functions - 3.6 Exercises - Page 223: 28


$f'(x) = \frac{1}{2x~\sqrt{2+ln~x}}$ The domain is $~~(\frac{1}{e^2}, \infty)$

Work Step by Step

$f(x) = \sqrt{2+ln~x}$ We can differentiate $f(x)$: $f'(x) = \frac{1}{2}(2+ln~x)^{-1/2}\cdot \frac{d}{dx}(2+ln~x)$ $f'(x) = \frac{1}{2~\sqrt{2+ln~x}}\cdot \frac{1}{x}$ $f'(x) = \frac{1}{2x~\sqrt{2+ln~x}}$ When we consider $~~\sqrt{2+ln~x}~~$ it is required that $~~2+ln~x \gt 0$ Then $~~ln~x \gt -2$ Then $~~x \gt e^{-2}$ Then $~~x \gt \frac{1}{e^2}$ The domain is $~~(\frac{1}{e^2}, \infty)$
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