## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises: 9

#### Answer

$H'(u)=2u-1$

#### Work Step by Step

$H(u)=(u-\sqrt{u})(u+\sqrt{u})$ Using the difference of two squares factorization formula, which is: $a^{2}-b^{2}=(a-b)(a+b)$ We can write the function like this: $H(u)=(u-\sqrt{u})(u+\sqrt{u})=u^{2}-(\sqrt{u})^{2}=u^{2}-u$ Now, differentiate each term: $H'(u)=2u-1$

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