Calculus 10th Edition

Published by Brooks Cole

Chapter 2 - Differentiation - Review Exercises: 31

Answer

$h'(x)=\dfrac{\sin{x}+2x\cos{x}}{2\sqrt{x}}.$

Work Step by Step

Product Rule $(h’(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$ $u(x)=\sqrt{x} ;u’(x)=\dfrac{1}{2\sqrt{x}}$ $v(x)=\sin{x} ;v’(x)=\cos{x}$ $h'(x)=(\dfrac{1}{2\sqrt{x}})(\sin{x})+(\cos{x})(\sqrt{x})$ $=\dfrac{\sin{x}+2x\cos{x}}{2\sqrt{x}}$

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