Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.5 The Chain Rule - 2.5 Exercises: 50

Answer

$$y'=\frac{2x+4}{(x+1)^{3/2}}$$ $$y''=\frac{-x-4}{(x+1)^{5/2}}$$

Work Step by Step

$$y'=\left(\frac{4x}{\sqrt{x+1}}\right)'=\frac{(4x)'\sqrt{x+1}-4x(\sqrt{x+1})'}{(\sqrt{x+1})^2}= \frac{4\sqrt{x+1}-4x\frac{1}{2\sqrt{x+1}}(x+1)'}{x+1}= \frac{4(x+1)-2x\cdot1}{(x+1)^{3/2}}=\frac{2x+4}{(x+1)^{3/2}}$$ $$y''=(y')'=\left(\frac{2x+4}{(x+1)^{3/2}}\right)'=\frac{(2x+4)'(x+1)^{3/2}-(2x+4)((x+1)^{3/2})'}{(x+1)^3}=\frac{2(x+1)^{3/2}-(2x+4)\frac{3}{2}(x+1)^{1/2}(x+1)'}{(x+1)^3}= \frac{2(x+1)^{3/2}-(x+2)\cdot3(x+1)^{1/2}\cdot1}{(x+1)^3}= \frac{2(x+1)-3(x+2)}{(x+1)^{5/2}}=\frac{2x+2-3x-6}{(x+1)^{5/2}}=\frac{-x-4}{(x+1)^{5/2}}$$
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