University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 159: 80


For $x=-1$, $$(fog)'=1$$

Work Step by Step

$f(u)=1-\frac{1}{u}$, $u=g(x)=\frac{1}{1-x}$, $x=-1$ Applying the Chain Rule: $$(fog)'=f'(g(x))g'(x)=f'(u)u'$$ We have $f'(u)=(1-\frac{1}{u})'=0-\Big(\frac{-1(u)'}{u^2}\Big)=\frac{1}{u^2}$ $u'=\Big(\frac{1}{1-x}\Big)'=\frac{-1(1-x)'}{(1-x)^2}=\frac{-1(-1)}{(1-x)^2}=\frac{1}{(1-x)^2}$ Therefore, $$(fog)'=\frac{1}{u^2}\times\frac{1}{(1-x)^2}=\frac{1}{\frac{1}{(1-x)^2}}\times\frac{1}{(1-x)^2}$$ $$=(1-x)^2\times\frac{1}{(1-x)^2}=1$$ For $x=-1$, $$(fog)'=1$$
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