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: 18

Answer

The Derivative is: $y'=\frac{-3x^2-8x-4}{(3x^2+x)^2}$

Work Step by Step

$y=\frac{4 - 3x}{3x^2 + x}$ Using the Quocient Rule: $y'=\frac{f'(x)⋅g(x)-f(x)⋅g'(x)}{g^2(x)}$ $y'=\frac{(0-(3)x^{1-1})(3x^2+x) - (4 - 3x)((2)3x^{2-1}+(1)x^{1-1})}{(3x^2+x)^2}$ $y'=\frac{-3(3x^2+x) - (4-3x)(2x+1)}{(3x^2+x)^2}$ $y'=\frac{-9x^2-3x-8x-4+6x^2+3x}{(3x^2+x)^2}$ $y'=\frac{-3x^2-8x-4}{(3x^2+x)^2}$
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