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 - Page 125: 13

Answer

(a): Using the Product Rule - $y'=-5x^{4}+12x^{2}-2x-3$ (b): Multiplying the factors to produce a sum of simpler terms to differentiate - $y'=-5x^{4}+12x^{2}-2x-3$

Work Step by Step

$y = (3 - x^{2}) (x^{3} - x + 1)$ (a) applying the Product Rule: $y'=f(x)\cdot\:g'(x) + g(x)\cdot\:f'(x)$ $y'=(3-x^{2})((3)x^{3-1}-(1)x^{1-1}+0) + (x^{3}-x+1)(0-(2)x^{2-1})$ $y'=(3-x^{2})(3x^{2}-1)+(x^{3}-x+1)(-2x)$ $y'=9x^{2}-3-3x^{4}+x^{2}-2x^{4}+2x^{2}-2x$ $y'=-5x^{4}+12x^{2}-2x-3$ (b) multiplying the factors to produce a sum of simpler terms to differentiate: $y = (3 - x^{2}) (x^{3} - x + 1)$ $y=3x^{3}+3(-x)+3\cdot \:1-x^2x^3-x^2(-x)-x^2\cdot \:1$ $y=3x^{3}-3x+3-x^{5}+x^{3}-x^{2}$ $y=-x^5+4x^3-x^2-3x+3$ Derivating the function using the Power Rule $y'=-(5)x^{5-1}+(3)(4)x^{3-1}-(2)x^{2-1}-(1)(3)x^{1-1}+(0)$ $y'=-5x^{4}+12x^{2}-2x-3$
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