The quotient rule states that if #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)#
