Answer
$y'=4x\cos(x)+(2-x^2)\sin(x)$.
Work Step by Step
$y=f(x)+g(x)\rightarrow f(x)=2x\sin(x)$; $g(x)=x^2cos(x)$
Using Product Rule:
$f'(x)=((u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$
$u(x)=2x ;u’(x)=2 $
$v(x)=\sin(x) ;v’(x)=\cos(x) $
$f'(x)=2x\cos(x)+2\sin(x)$.
Using Product Rule:
$g'(x)=((u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$
$u(x)=x^2 ;u’(x)=2x $
$v(x)=\cos(x) ;v’(x)=-\sin(x) $
$g'(x)=(2x)(\cos(x))-x^2\sin(x)$.
Using Sum Rule:
$y'=f'(x)+g'(x)=4x\cos(x)+(2-x^2)\sin(x)$.
