Calculus 10th Edition

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 - 5.1 Exercises: 60

Answer

$=\frac{1}{\sqrt{x^2+4}}$

Work Step by Step

$f(x)=ln(x+\sqrt{4+x^2})$ $=ln(x+(4+x^2)^{\frac{1}{2}})$ let u = $x+(4+x^2)^{\frac{1}{2}}$ $u' = 1+ ( \frac{1}{2}(4+x^2)^{-\frac{1}{2}})(2x)$ $= 1+ ( (4+x^2)^{-\frac{1}{2}})x$ $f'(x)=\frac{u'}{u}$ $=\frac{1+ ( (4+x^2)^{-\frac{1}{2}})x}{x+(4+x^2)^{\frac{1}{2}}}$ $=\frac{1}{\sqrt{x^2+4}}$
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