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