If f(x)=x2sinxf(x) = x^2 \sin xf(x)=x2sinx, then f′(x)=f'(x) =f′(x)=Ax2cosxx^2 \cos xx2cosxB2xsinx2x \sin x2xsinxCx2cosx+2xsinxx^2 \cos x + 2x \sin xx2cosx+2xsinxcheck_circleD2xcosx2x \cos x2xcosxExplanation(x2)′sinx+x2(sinx)′=2xsinx+x2cosx(x^2)' \sin x + x^2 (\sin x)' = 2x \sin x + x^2 \cos x(x2)′sinx+x2(sinx)′=2xsinx+x2cosx.