Answer
$y’=\frac{e^{\sqrt{x}}} {2\sqrt{x}}$
Work Step by Step
$y=e^{\sqrt{x}}$
let u=$\sqrt{x}$
u'=$\frac{1}{2}x^{-\frac{1}{2}}$
$\frac{d}{dx}e^u=u'e^u$
$y’=\frac{1}{2}x^{-\frac{1}{2}}e^{\sqrt{x}}$
$=\frac{e^{\sqrt{x}}} {2\sqrt{x}}$
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 - 5.4 Exercises - Page 352: 35
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