## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises: 19

#### Answer

$y'=-\dfrac{1}{s^{2}}+\dfrac{3}{2\sqrt{s^{5}}}$

#### Work Step by Step

$y=\dfrac{s-\sqrt{s}}{s^{2}}$ Rewrite the function like this: $y=s^{-2}(s-s^{1/2})=s^{-1}-s^{-3/2}$ Differentiate each term: $y'=-s^{-2}-(-\dfrac{3}{2})s^{-5/2}=-s^{-2}+\dfrac{3}{2}s^{-5/2}=-\dfrac{1}{s^{2}}+\dfrac{3}{2s^{5/2}}=-\dfrac{1}{s^{2}}+\dfrac{3}{2\sqrt{s^{5}}}$ Please note that there was an easier way to differentiate the function than using the quotient rule

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