Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.1 - Derivatives of Polynomials and Exponential Functions - 3.1 Exercises - Page 180: 27



Work Step by Step

$$G(q)=(1+q^{-1})^2$$ $$G(q)=1+(2\times1\times q^{-1})+(q^{-1})^2$$ $$G(q)=1+2q^{-1}+q^{-2}$$ So, $$G'(q)=\frac{d}{dq}(1)+2\frac{d}{dq}(q^{-1})+\frac{d}{dq}q^{-2}$$ $$G'(q)=0+2\times(-1)\times q^{-2}+(-2)\times q^{-3}$$ $$G'(q)=-2\times\frac{1}{q^2}-2\times\frac{1}{q^3}$$ $$G'(q)=\frac{-2}{q^2}-\frac{2}{q^3}$$
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