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: 14

Answer

(a) applying the Product Rule: $y'=30x^{2}+14x-12$ (b) multiplying the factors to produce a sum of simpler terms to differentiate: $y'=30x^{2}+14x-12$

Work Step by Step

$y=(2x+3)(5x^2-4x)$ (a) applying the Product Rule: $y'=f'(x)⋅g(x)+f(x)⋅g'(x)$ $y'=((1)(2)x^{1-1}+0)(5x^2-4x)+(2x+3)((2)(5)x^{2-1}-(1)(4)x^{1-1})$ $y'=(2)(5x^2-4x)+(2x+3)(10x-4)$ $y'=10x^2-8x+20x^2-8x+30x-12$ $y'=30x^2+14x-12$ (b) multiplying the factors to produce a sum of simpler terms to differentiate: $y=(2x+3)(5x^2-4x)$ $y=10x^3-8x^2+15x^2-12x$ $y=10x^3+7x^2-12x$ Derivating the function using the Power Rule $y'=(3)(10)x^{3-1}+(2)(7)x^{2-1}-(1)(12)x^{1-1}$ $y'=30x^{2}+14x-12$
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