Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - Review - Exercises: 15

Answer

$\dfrac {3}{2}\sqrt {x}-\dfrac {1}{2\sqrt {x}}-\dfrac {1}{x\sqrt {x}}$

Work Step by Step

$y=\dfrac {x^{2}-x+2}{\sqrt {x}}\Rightarrow y'=\dfrac {\dfrac {d}{dx}\left( x^{2}-x+2\right) \times \sqrt {x}-\dfrac {d}{dx}\left( \sqrt {x}\right) \left( x^{2}-x+2\right) }{\left( \sqrt {x}\right) ^{2}}=\dfrac {\left( 2x-1\right) \sqrt {x}-\dfrac {1}{2}\dfrac {\left( x^{2}-x+2\right) }{\sqrt {x}}}{x}=\dfrac {3}{2}\sqrt {x}-\dfrac {1}{2\sqrt {x}}-\dfrac {1}{x\sqrt {x}}$
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