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 - Page 425: 8



Work Step by Step

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