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 163: 31


$$ y'= \frac{-x^2 - 2x+1}{(x^2+1)^2}.$$

Work Step by Step

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