Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.2 Derivatives And Integrals Involving Logarithmic Functions - Exercises Set 6.2: 7

Answer

$y'=\dfrac{1-x^2}{x+x^3}$

Work Step by Step

In order to derivate this function you have to apply the chain rule Let's make a «u» substitution to make it easier $f(u) = $ $u = \dfrac{x}{1+x^2}$ Derivate the function: $f'(u) = \dfrac{u'}{u}$ Now let's find u' Note: Here you have to apply quotient rule $u' = \dfrac{1-x^2}{(1+x^2)^2}$ Then undo the substitution, simplify and get the answer: $f'(u) = \dfrac{1+x^2}{x} \times \dfrac{1-x^2}{(1+x^2)^2}$ $f'(u) = \dfrac{1-x^2}{x(1+x^2)}$ $f'(u)=\dfrac{1-x^2}{x+x^3}$
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