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: 91

Answer

$\displaystyle \frac{dy}{dx}=\frac{3x^{3}+15x^{2}-8x}{2(x+1)^{3}\sqrt{3x-2}}$

Work Step by Step

$y=\displaystyle \frac{x^{2}\sqrt{3x-2}}{(x+1)^{2}}$ $...$apply ln( ) to both sides ... on the RHS, apply$ \displaystyle \ln\frac{M}{N}=\ln M-\ln M$ $\ln y=\ln(x^{2}(3x-2)^{1/2})-(x+1)^{2}$ ... now, apply $\ln(M\cdot N)=\ln M+\ln N$ and $\ln M^{n}=n\ln M$ $\displaystyle \ln y=2\ln x+\frac{1}{2}\ln(3x-2)-2\ln(x+1)$ $\displaystyle \frac{1}{y}(\frac{dy}{dx})=\frac{2}{x}+\frac{3}{2(3x-2)}-\frac{2}{x+1}$ $\displaystyle \frac{1}{y}(\frac{dy}{dx})=\frac{2\cdot 2(3x-2)(x+1)+3\cdot x(x+1)-2\cdot 2x(3x-2)}{2x(3x-2)(x+1)}$ $\displaystyle \frac{1}{y}(\frac{dy}{dx})=\frac{4(3x^{2} +x-2)+3x^{2}+3x-12x^{2}+8x}{2x(3x-2)(x+1)}\qquad.../\times y$ $\displaystyle \frac{dy}{dx}=y[\frac{3x^{2}+15x-8}{2x(3x-2)(x+1)}]\quad$ ... insert expression for y $\displaystyle \frac{dy}{dx}=\frac{x^{2}(3x-2)^{1/2}}{(x+1)^{2}}\cdot\frac{3x^{2}+15x-8}{2x(3x-2)(x+1)}$ $\displaystyle \frac{dy}{dx}=\frac{3x^{3}+15x^{2}-8x}{2(x+1)^{3}(3x-2)^{1/2}}$ $\displaystyle \frac{dy}{dx}=\frac{3x^{3}+15x^{2}-8x}{2(x+1)^{3}\sqrt{3x-2}}$
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