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 - Page 158: 28



Work Step by Step

$$F'(t)=\left(\frac{t^2}{\sqrt{t^3+1}}\right)'=\frac{(t^2)'\sqrt{t^3+1}-t^2(\sqrt{t^3+1)'}}{(\sqrt{t^3+1})^2}=\frac{2t\sqrt{t^3+1}-t^2\frac{1}{2\sqrt{t^3+1}}(t^3+1)'}{t^3+1}=\frac{4t(t^3+1)-t^2\cdot3t^3}{2(t^3+1)^{3/2}}=\frac{4t^4+4t-3t^4}{2(t^3+1)^{3/2}}= \frac{t(t^3+4)}{2(t^3+1)^{3/2}}$$
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