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: 9

Answer

First Derivative: $y'=12x-10+10x^{-3}$ Second Derivative: $y''=12-30x^{-4}$

Work Step by Step

Using the Power Rule: First Derivative: $y=6x^{2}-10x-5x^{-2}$ $y'=(2)(6)x^{2-1}-(1)(10)x^{1-1}-(-2)(5)x^{-2-1}$ $y'=12x-10+10x^{-3}$ Second Derivative $y'=12x-10+10x^{-3}$ $y''=(1)(12)x^{1-1}-(0)+(-3)(10)x^{-3-1}$ $y''=12-30x^{-4}$ (the derivative of 10 will be 0 because of the rule of the constant function)
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