Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.2 - Natural Logarithms - Exercises 7.2 - Page 381: 68


$ y^{\prime}=\frac{1}{3} \sqrt[3]{\frac{x(x+1)(x-2)}{\left(x^{2}+1\right)(2 x+3)}} (\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x-2}-\frac{2 x}{x^{2}+1}-\frac{2}{2 x+3}) $

Work Step by Step

Given $$ y=\sqrt[3]{\frac{x(x+1)(x-2)}{\left(x^{2}+1\right)(2 x+3)}}$$ So, we have \begin{aligned} & \Rightarrow \ln y=\ln \sqrt[3]{\frac{x(x+1)(x-2)}{\left(x^{2}+1\right)(2 x+3)}}\\ & \Rightarrow \ln y=\frac{1}{3}\ln \frac{x(x+1)(x-2)}{\left(x^{2}+1\right)(2 x+3)}\\ & \Rightarrow \ln y=\frac{1}{3} [\ln[ x (x+1) (x-2)]-\ln [\left(x^{2}+1\right) (2 x+3)] ]\\ & \Rightarrow \ln y=\frac{1}{3} [\ln x+\ln (x+1)+\ln (x-2)-\ln \left(x^{2}+1\right)-\ln (2 x+3) ]\\ &\text{differentiate both sides with respect to } x\\ &\Rightarrow y^{\prime}=\frac{y}{3} (\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x-2}-\frac{2 x}{x^{2}+1}-\frac{2}{2 x+3}) \\ &\Rightarrow y^{\prime}=\frac{1}{3} \sqrt[3]{\frac{x(x+1)(x-2)}{\left(x^{2}+1\right)(2 x+3)}} (\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x-2}-\frac{2 x}{x^{2}+1}-\frac{2}{2 x+3}) \\ \end{aligned}
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