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 - Page 125: 25


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

Work Step by Step

Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=(4-3x-x^2); u'(x)=(-2x-3)$ $v(x)=(x^2-1); v'(x)=2x$ $f'(x)=\frac{(-2x-3)(x^2-1)-(2x)(4-3x-x^2)}{(x^2-1)^2}=$ $\frac{(-2x^3-3x^2+2x+3)+(-8x+6x^2+2x^3)}{(x+1)^2(x-1)^2}=$ $\frac{3(x^2-2x+1)}{(x-1)^2(x+1)^2}=\frac{3(x-1)^2}{(x-1)^2(x+1)^2}= \frac{3}{(x+1)^2}$
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