Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.4 Exercises: 31

Answer

$f'(v)=\dfrac{-9(1-2v)^2}{(1+v)^4}.$

Work Step by Step

Using the Quotient Rule and the Chain Rule: $f'(v)=(\dfrac{y(v)}{z(v)})'=\dfrac{y'(v)z(v)-z'(v)y(v)}{(z(v))^2}$ $y(v)=(1-2v)^3; y'(v)=(3)(-2)(1-2v)^{3-1}=-6(1-2v)^2.$ $z(v)=(1+v)^3; z'(v)=(3)(1)(1+v)^{3-1}=3(1+v)^2$ $f'(v)=\dfrac{-6(1-2v)^2(1+v)^3-3(1+v)^2(1-2v)^3}{((1+v)^3)^2}$ $=\dfrac{-9(1-2v)^2}{(1+v)^4}$
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