Calculus (3rd Edition)

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

Chapter 3 - Differentiation - 3.7 The Chain Rule - Exercises - Page 146: 54



Work Step by Step

Since $ y=\frac{(x+1)^{1/2}}{x+2}$, then we can rewrite it as follows $$ y=(x+1)^{1/2} (x+2)^{-1}.$$ Now, by using the product rule, the derivative $ y'$ is given by $$ y'=\frac{1}{2}(x+1)^{-1/2} (x+2)^{-1}-(x+1)^{1/2} (x+2)^{-2}\\ =\frac{1}{2(x+2)\sqrt{x+1}}-\frac{\sqrt{x+1}}{(x+2)^2}\\ =\frac{x+2-2(x+1)}{(x+2)^2\sqrt{x+1}}=\frac{-x}{(x+2)^2\sqrt{x+1}}.$$
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