Calculus 10th Edition

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

Chapter 2 - Differentiation - Review Exercises: 40

Answer

$g'(x)=(3-x^2)\sin{x}+5x\cos{x}$

Work Step by Step

$g(x)=f(x)+h(x)\rightarrow f(x)=3x\sin{x}$ ; $h(x)=x^2\cos{x}$ Product Rule $(f’(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$ $u(x)=3x ;u’(x)=3 $ $v(x)=\sin{x} ;v’(x)=\cos{x} $ $f'(x)=(3)(\sin{x})+(3x)(\cos{x})=3(\sin{x}+x\cos{x}).$ Product Rule $(h’(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$ $u(x)=x^2 ;u’(x)=2x $ $v(x)=\cos{x} ;v’(x)=-\sin{x} $ $h'(x)=(2x)(\cos{x})+(x^2)(-\sin{x})=x(2\cos{x}-x\sin{x})$ $g'(x)=h'(x)+f'(x)=x(2\cos{x}-x\sin{x})+3(\sin{x}+x\cos{x})$ $=(3-x^2)\sin{x}+5x\cos{x}.$
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