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: 28

Answer

$$F'(z)=-\frac{1}{z^2}(\frac{2A}{z}+B)$$

Work Step by Step

$$F(z)=\frac{A+Bz+Cz^2}{z^2}$$ $$F(z)=\frac{A}{z^2}+\frac{B}{z}+C$$ $$F(z)=Az^{-2}+Bz^{-1}+C$$ Therefore, the derivative of $F(z)$ is $$F'(z)=A\frac{d}{dz}(z^{-2})+B\frac{d}{dz}(z^{-1})+\frac{d}{dz}(C)$$ $$F'(z)=A\times(-2)z^{-3}+B\times(-1)z^{-2}+0$$ $$F'(z)=-2Az^{-3}-Bz^{-2}$$ $$F'(z)=-z^{-2}(2Az^{-1}+B)$$ $$F'(z)=-\frac{1}{z^2}(\frac{2A}{z}+B)$$
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