find f'''(x) if f(x)=xsinx
is it just simply f'''(x)=-xcosx
$\displaystyle f(x)=x\sin x$
$\displaystyle f'(x)= (x)'\cdot \sin x+x\cdot (\sin x)'$
$\displaystyle =1\cdot \sin x+x\cdot \cos x$
$\displaystyle =\sin x+x\cos x$
$\displaystyle f''(x)=(\sin x)'+(x)'\cos x+x\cdot (\cos x)'$
$\displaystyle =\cos x+\cos x-x\sin x=2\cos x-x\sin x$
$\displaystyle f'''(x)=(2\cos x)'+(-x)'\sin x +(-x)(\sin x)'$
$\displaystyle =-2\sin x-\sin x-x\cos x=-3\sin x-x\cos x$