Calculus 10th Edition

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

Chapter 2 - Differentiation - Review Exercises: 33

Answer

$f'(x)=\dfrac{-1-x^2}{(x^2-1)^2}.$

Work Step by Step

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