Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.1 Exercises: 44

Answer

$\displaystyle \frac{4x}{2x^{2}+1}$

Work Step by Step

By theorem 5.3/2: $\displaystyle \frac{d}{dx}[\ln u]=\frac{1}{u}\cdot\frac{du}{dx}=\frac{u^{\prime}}{u},\ \ \ (u>0)$ Here, u(x)$=2x^{2}+1,\displaystyle \ \ \ \frac{du}{dx}=4x$ $\displaystyle \frac{d}{dx}[\ln(2x^{2}+1)]=\frac{1}{2x^{2}+1}\cdot 4x=\frac{4x}{2x^{2}+1}$
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