University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.3 - Differentiation Rules - Exercises - Page 136: 45

Answer

$$y'=\frac{2x^3-7}{x^2}$$ $$y''=\frac{2x^3+14}{x^3}$$

Work Step by Step

$$y=\frac{x^3+7}{x}$$ 1) First derivative: Apply the Derivative Quotient Rule: $$y'=\frac{x(x^3+7)'-(x)'(x^3+7)}{x^2}=\frac{x(2x^2+0)-1\times(x^3+7)}{x^2}$$ $$y'=\frac{3x^3-x^3-7}{x^2}=\frac{2x^3-7}{x^2}$$ 2) Second derivative: Apply the Derivative Quotient Rule: $$y''=\frac{x^2(2x^3-7)'-(x^2)'(2x^3-7)}{x^4}=\frac{x^2(6x^2-0)-2x(2x^3-7)}{x^4}$$ $$y''=\frac{6x^4-4x^4+14x}{x^4}=\frac{2x^4+14x}{x^4}=\frac{2x^3+14}{x^3}$$
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