Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 386: 53


$$y'=\frac{(x+1)^3}{(4x-2)^2}\left(\frac{3}{x+1}-\frac{4}{2x-1}\right) .$$

Work Step by Step

Since $ y=\frac{(x+1)^3}{(4x-2)^2}$, then by taking the $\ln $ for both sides, we have $$\ln y=\ln\left( \frac{(x+1)^3}{(4x-2)^2}\right)=\ln(x+1)^3-\ln (4x-2)^2\\ =3\ln(x+1)-2\ln (4x-2).$$ Now, taking the derivative, we get $$\frac{y'}{y}=\frac{3}{x+1}-\frac{8}{4x-2}$$ Hence $$ y'=y\left(\frac{3}{x+1}-\frac{8}{4x-2}\right)\\ y'=\frac{(x+1)^3}{(4x-2)^2}\left(\frac{3}{x+1}-\frac{4}{2x-1}\right) .$$
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