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: 7

Answer

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

Work Step by Step

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