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: 90

Answer

$\displaystyle \frac{dy}{dx}=\frac{4x^{2}+9x+4}{2\sqrt{(x+1)(x+2)}}$

Work Step by Step

$y=\sqrt{x^{2}(x+1)(x+2)}$ $...$apply ln( ) to both sides ... on the RHS, apply $\ln M^{n}=n\ln M,$ $\displaystyle \ln y=\frac{1}{2}\ln[x^{2}(x+1)(x+2)]$ ... on the RHS, apply $\ln(M\cdot N)=\ln M+\ln N$ $\displaystyle \ln y=\frac{1}{2}[2\ln(x)+\ln(x+1)+\ln(x+2)]\qquad $...$/\displaystyle \frac{d}{dx}$ $\displaystyle \frac{1}{y}(\frac{dy}{dx})=\frac{1}{2}[\frac{2}{x}+\frac{1}{x+1}+\frac{1}{x+2}]$ $\displaystyle \frac{1}{y}(\frac{dy}{dx})=\frac{1}{2}[\frac{2(x+1)(x+2)+x(x+2)+x(x+1)}{x(x+1)(x+2)}]$ $\displaystyle \frac{1}{y}(\frac{dy}{dx})=\frac{1}{2}[\frac{2(x^{2}+3x+2)+x^{2}+2x+x^{2}+x}{x(x+1)(x+2)}]$ $\displaystyle \frac{1}{y}(\frac{dy}{dx})=\frac{4x^{2}+9x+4}{2x(x+1)(x+2)} \qquad/\times y$ $... y=|x|(x+1)^{1/2}(x+2)^{1/2}=x(x+1)^{1/2}(x+2)^{1/2}$, since $x > 0$ $\displaystyle \frac{dy}{dx}=\frac{(4x^{2}+9x+4)\cdot y}{2x(x+1)(x+2)}$ $\displaystyle \frac{dy}{dx}=\frac{(4x^{2}+9x+4)x(x+1)^{1/2}(x+2)^{1/2}}{2x(x+1)(x+2)}$ $\displaystyle \frac{dy}{dx}=\frac{4x^{2}+9x+4}{2(x+1)^{1/2}(x+2)^{1/2}}$ $\displaystyle \frac{dy}{dx}=\frac{4x^{2}+9x+4}{2\sqrt{(x+1)(x+2)}}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.