Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.3 Differentiation Rules - Exercises 3.3: 17

Answer

The Derivative is: $y'=-\frac{19}{(3x-2)^2}$

Work Step by Step

$y=\frac{2x+5}{3x-2}$ Using the Quocient Rule: $y'=\frac{f'(x)⋅g(x)-f(x)⋅g'(x)}{g^2(x)}$ $y'=\frac{((1)2x^{1-1}+0)⋅(3x-2)-(2x+5)⋅((1)3x^{1-1}-0)}{(3x-2)^{2}}$ $y'=\frac{2(3x-2)-3(2x+5)}{(3x-2)^2}$ $y'=\frac{6x-4-6x-15}{(3x-2)^2}$ $y'=-\frac{19}{(3x-2)^2}$
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