Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.3 Exercises: 29

Answer

$f'(x)=\frac{3x+1}{2x\sqrt{x}}$

Work Step by Step

Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=3x-1; u'(x)=3$ $v(x)=\sqrt{x}; v'(x)=\frac{1}{2\sqrt{x}}$ $f'(x)=\frac{(3)(\sqrt{x})-(\frac{1}{2\sqrt{x}})(3x-1)}{(\sqrt{x})^2}$ $=\frac{(3)(\frac{2x}{2\sqrt{x}})-(\frac{1}{2\sqrt{x}})(3x-1)}{(\sqrt{x})^2}$ $=\frac{6x-3x+1}{2(\sqrt{x})(x)}=\frac{3x+1}{2x\sqrt{x}}$
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