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: 27

Answer

$f'(x)=\frac{x^2+6x-3}{(x+3)^2}$

Work Step by Step

$f(x)=x(\frac{(x+3)-4}{(x+3)})=x(\frac{x-1}{x+3})=\frac{x^2-x}{x+3}$ Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=x^2-x; u'(x)=2x-1$ $v(x)=x+3; v'(x)=1$ $f'(x)=\frac{(2x-1)(x+3)-(1)(x^2-x)}{(x+3)^2}$ $=\frac{2x^2+5x-3-x^2+x}{(x+3)^2}=\frac{x^2+6x-3}{(x+3)^2}$
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