Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.3 Introduction To Techniques Of Differentiation - Exercises Set 2.3 - Page 140: 22

Answer

$y'(1) = 0$

Work Step by Step

$y=\dfrac{1+x+x^2+x^3+x^4+x^5+x^6}{x^3}$ If we divide each term in the numerator by $x^3$, we have: $y = x^{-3}+x^{-2}+x^{-1}+1+x+x^2+x^3$ Taking the derivative, $\dfrac{dy}{dx}= (-3)x^{-3-1}+(-2)x^{-2-1}+(-1)x^{-1-1}+0+x^{1-1}+(2)x^{2-1}+(3)x^{3-1} = -3x^{-4}-2x^{-3}-x^{-2}+1+2x+3x^2$ Plugging in $x=1$, $\dfrac{dy}{dx}$ at $x =1$ is $-3-2-1+1+2+3 = \boxed{0}$
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