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: 10

Answer

The derivative is $\frac{3x\sqrt x+2x}{(2\sqrt x +1)^2}.$

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} $ $u(x)=x^2; u'(x)=2x$ $v(x)=2\sqrt x +1; v'(x)=\frac{1}{\sqrt x}$ $f'(x)=\dfrac{(2x)(2\sqrt x+1)-(x^2)(\frac{1}{\sqrt x})}{(2\sqrt x+1)^2}$ $=\dfrac{4x\sqrt x+2x-x\sqrt x}{(2\sqrt x +1)^2}=\dfrac{3x\sqrt x+2x}{(2\sqrt x +1)^2}$.
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