Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.3 Product and Quotient Rules - Exercises - Page 122: 35


$$ g'(z)= 2z-1.$$

Work Step by Step

For the sake of simplicity, we rewrite $ g(z)$ as follows $$ g(z)=\frac{(z+2)(z-2)}{(z-1)} \, \frac{(z-1)(z+1)}{z+2}\\ =(z-2)(z+1).$$ Using the product rule, the derivative $ g'(z)$ is given by $$ g'(z)= (z+1)+(z-2)=2z-1.$$
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