Answer
$f''(x)=2\cos{x}-x\sin{x}$
Work Step by Step
First Derivative:
Product Rule $(f’(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$
$u(x)=x ;u’(x)=1 $
$v(x)=\sin{x} ;v’(x)=\cos{x} $
$f'(x)=x\cos{x}+\sin{x}$
Second Derivative:
$f'(x)=g(x)+h(x)\rightarrow g(x)=\sin{x}$; $h(x)=x\cos{x}$
Product Rule: $h'(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x)$
$u(x)=x ;u’(x)=1 $
$v(x)=\cos{x} ;v’(x)=-\sin{x} $
$h'(x)=\cos{x}-x\sin{x}.$
$g'(x)=\frac{d}{dx}\sin{x}=\cos{x}.$
$f''(x)=g'(x)+h'(x)=2\cos{x}-x\sin{x}.$
