The quotient rule states that if f(x)=(g(x))/(h(x))f(x)=g(x)h(x) then f'(x)=(g'(x)h(x)-g(x)h'(x))/(h^2(x)). In this case, g(x)=x-sinx and h(x)=x^2cosx.
To find g'(x) we need to use the difference rule, which states that if g(x)=u(x)-v(x) then g'(x)=u'(x)-v'(x). Here, u(x)=x and v(x)=sinx, so u'(x)=d/dxx=1 and v'(x)=d/dxsinx=cosx, therefore g'(x)=1-cosx.
To find h'(x) we need to use the product rule, which states that if h(x)=u(x)v(x) then h'(x)=u'(x)v(x)+u(x)v'(x). Here, u(x)=x^2 and v(x)=cosx, so u'(x)=d/dxx^2=2x and v'(x)=d/dxcosx="-"sinx, therefore h'(x)=2xcosx-x^2sinx.
h^2(x)=(x^2cosx)^2=x^4cos^2x
Putting everything back together gives f'(x)=((1-cosx)(x^2cosx)-(x-sinx)(2xcosx-x^2sinx))/(x^4cos^2x), which simplifies to ((x^2cosx-x^2cos^2x)-(2x^2cosx-x^3sinx-2xsinxcosx+x^2sin^2x))/(x^4cos^2x), which simplifies to (x^2cosx-x^2cos^2x-2x^2cosx+x^3sinx+2xsinxcosx-x^2sin^2x)/(x^4cos^2x), which simplifies to (x^3sinx-x^2cosx-x^2sin^2x-x^2cos^2x+2xsinxcosx)/(x^4cos^2x), which factors to (x(x^2sinx-xcosx-x(sin^2x+cos^2x)+2sinxcosx))/(x^4cos^2x).
Since sin2theta=2sinthetacostheta and sin^2theta+cos^2theta=1, this simplifies to (x^2sinx-xcosx-x+sin2x)/(x^3cos^2x)
