Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises: 25

Answer

$f'(x)=\dfrac{2cx}{(x^{2}+c)^{2}}$

Work Step by Step

$f(x)=\dfrac{x}{x+\dfrac{c}{x}}$ (Here $c$ is a constant) Let's use algebra to make some changes to the function: $f(x)=\dfrac{x}{x+\dfrac{c}{x}}=\dfrac{x}{\dfrac{x^{2}+c}{x}}=\dfrac{x^{2}}{x^{2}+c}$ Differentiate using the quotient rule: $f'(x)=\dfrac{(x^{2}+c)(x^{2})'-(x^{2})(x^{2}+c)'}{(x^{2}+c)^{2}}=\dfrac{(x^{2}+c)(2x)-(x^{2})(2x)}{(x^{2}+c)^{2}}$ Simplify $...=\dfrac{2x^{3}+2cx-2x^{3}}{(x^{2}+c)^{2}}=\dfrac{2cx}{(x^{2}+c)^{2}}$
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