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 - Page 326: 89


$\displaystyle \frac{dy}{dx}=\frac{2x^{2}+1}{(x^{2}+1)^{1/2}}$

Work Step by Step

$y=x\sqrt{x^{2}+1}$ $y=x\cdot(x^{2}+1)^{1/2}$ $...$apply ln( ) to both sides ... on the RHS, $\ln(M\cdot N)=\ln M+\ln N$ and $\ln N^{r}=r\ln N$ $\displaystyle \ln y=\ln x+\frac{1}{2}\ln(x^{2}$+$1)\qquad $...$/\displaystyle \frac{d}{dx}$ ... chain rule for the LHS and second term on the RHS $\displaystyle \frac{1}{y}(\frac{dy}{dx})=\frac{1}{x}+\frac{1}{2}\cdot\frac{1}{x^{2}+1}\cdot 2x$ $\displaystyle \frac{1}{y}(\frac{dy}{dx})=\frac{1}{x}+\frac{x}{x^{2}+1}$ $\displaystyle \frac{1}{y}(\frac{dy}{dx})=\frac{x^{2}+1+x^{2}}{x(x^{2}+1)} \qquad$...$/\times y$ $\displaystyle \frac{dy}{dx}=y[\frac{2x^{2}+1}{x(x^{2}+1)}]\qquad$ $... y=x\sqrt{x^{2}+1}=x(x^{2}$+$1)^{1/2}$ $\displaystyle \frac{dy}{dx}=\frac{x(x^{2}+1)^{1/2}(2x^{2}+1)}{x(x^{2}+1)}$ $\displaystyle \frac{dy}{dx}=\frac{2x^{2}+1}{(x^{2}+1)^{1/2}}$
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