Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.3 Differentiation Formulas - 2.3 Exercises: 39

Answer

$F'(x)=4x+1+\dfrac{12}{x^3}$

Work Step by Step

$F(x)=\dfrac{2x^5+x^4-6x}{x^3}$ Use the quotient rule: $\Bigg(\dfrac{f}{g}\Bigg)'=\dfrac{gf'-fg'}{g^2}$ so... $F'(x)=\dfrac{(x^3)(10x^4+4x^3-6)-(2x^5+x^4-6x)(3x^2)}{x^6}$ Distribute: $F'(x)=\dfrac{10x^7+4x^6-6x^3-6x^7-3x^6+18x^3}{x^6}$ Combine like terms: $F'(x)=\dfrac{4x^7+x^6+12x^3}{x^6}$ Simplify: $F'(x)=\dfrac{4x^7}{x^6}+\dfrac{x^6}{x^6}+\dfrac{12x^3}{x^6}$ $F'(x)=4x+1+\dfrac{12}{x^3}$
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