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: 54


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

Work Step by Step

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