Calculus (3rd Edition)

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

Chapter 3 - Differentiation - 3.7 The Chain Rule - Exercises - Page 146: 35


$f'(x) = -\dfrac{8}{(x-1)^2} \bigg (\dfrac{x+1}{x-1} \bigg )^3$

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 $u = \dfrac{x+1}{x-1}$ $f(u) = u^4 $ Derivate the function: $f'(u) = 4u^3u'$ Now let's find u' $u' = -\dfrac{2}{(x-1)^2}$ Then undo the substitution, simplify and get the answer: $f'(x) = 4 \bigg ( -\dfrac{2}{(x-1)^2}\bigg ) \bigg (\dfrac{x+1}{x-1} \bigg )^3$ $f'(x) = -\dfrac{8}{(x-1)^2} \bigg (\dfrac{x+1}{x-1} \bigg )^3$
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