Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises: 7

Answer

$y'=\dfrac{10}{(3-4x)^{2}}$

Work Step by Step

$g(x)=\dfrac{1+2x}{3-4x}$ Differentiate using the quotient rule $g'(x)=\dfrac{(3-4x)(1+2x)'-(1+2x)(3-4x)'}{(3-4x)^{2}}=\dfrac{(3-4x)(2)-(1+2x)(-4)}{(3-4x)^{2}}=...$ Evaluate the products in the numerator $...=\dfrac{6-8x+4+8x}{(3-4x)^{2}}=\dfrac{10}{(3-4x)^{2}}$ $y'=\dfrac{10}{(3-4x)^{2}}$
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