Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.3 Exercises: 30

Answer

$f'(x)=\frac{6+5\sqrt{x}}{6\sqrt[3]{x^2}}$

Work Step by Step

Product Rule $f'(x)=((u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$ $u(x)=\sqrt[3]{x} ;u’(x)=\frac{1}{3\sqrt[3]{x^2}} $ $v(x)=3+\sqrt{x} ;v’(x)=\frac{1}{2\sqrt{x}} $ $f'(x)=(\frac{1}{3\sqrt[3]{x^2}})(3+\sqrt{x})+(\frac{1}{2\sqrt{x}})(\sqrt[3]{x})$ $=\frac{3+\sqrt{x}}{3\sqrt[3]{x^2}}+\frac{1}{2\sqrt[6]{x}}=\frac{6+5\sqrt{x}}{6\sqrt[3]{x^2}}$
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