Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - Chapter Review Exercises - Page 164: 66


$$ y'= -\frac{8}{(x+1)^{3}} .$$

Work Step by Step

Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Recall that $(x^n)'=nx^{n-1}$ Given $ y=\frac{4x}{x+1}$, then, by using the quotient rule, we have $$ y'=\frac{(x+1)(4)-4x(1)}{(x+1)^2}=\frac{4}{(x+1)^2}=4(x+1)^{-2},$$ and the seond derivative $ y''$ is given by $$ y'=(-2)4(x+1)^{-3}=-8(x+1)^{-3}=-\frac{8}{(x+1)^{3}} .$$
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