Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 43


$$ y' =\frac{4x+11}{(x+1)(2x+9)}$$

Work Step by Step

Recall that $(\ln x)'=\dfrac{1}{x}$ Recall the product rule: $(uv)'=u'v+uv'$ Since $ y=\ln((x+1)(2x+9))$, by using the chain rule and the product rule, we have $$ y'=\frac{(x+1)'(2x+9)+(x+1)(2x+9)'}{(x+1)(2x+9)}\\ =\frac{(1)(2x+9)+(x+1)(2)}{(x+1)(2x+9)}=\frac{4x+11}{(x+1)(2x+9)}$$
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