Given:
#f(x)=(sin^2x+cosx)/(sinx-cos^2x)#
Let #u=sin^2x+cosx#
Then,
#(du)/dx=2sinxcosx-sinx#
Let #v=sinx-cos^2x#
Then,
#(dv)/dx=cosx-2cosx(-sinx)=cosx+2cosxsinx#
Rearranging
#(dv)/dx=2sinxcosx+cosx#
We have,
#d/dx(u/v)=(v(du)/dx-u(dv)/dx)/v^2#
Substituting
#d/dx((sin^2x+cosx)/(sinx-cos^2x))=((sinx-cos^2x)(2sinxcosx-sinx)-(sin^2x+cosx)(2sinxcosx+cosx))/(sinx-cos^2x)^2#
Simplifying
#((2sin^2xcosx-2sinxcos^3x-sin^2x+cos^2xsinx)-(2sin^3xcosx+2sinxcos^2x+sin^2xcosx+cos^2x))/(sinx-cos^2x)^2#
#((2sin^2xcosx-2sinxcos^3x-sin^2x+cos^2xsinx-2sin^3xcosx-2sinxcos^2x-sin^2xcosx-cos^2x))/(sinx-cos^2x)^2#
#(sin^2xcosx-cos^2xsinx-(sin^2x+cos^2x)-2sinx(cos^2x+sin^2x))/(sinx-cos^2x)^2#
Knowing that #cos^2x+sin^2x=1=sin^2x+cos^2x#
#(sin^2xcosx-cos^2xsinx-1-2sinx(1))/(sinx-cos^2x)^2#
#(sinxcosx(sinx-cosx)-1-2sinx)/(sinx-cos^2x)^2#
#(sinxcosx(sinx-cosx)-(1+2sinx))/(sinx-cos^2x)^2#
Thus, the derivative of
#(sin^2x+cosx)/(sinx-cos^2x)# is
#(sinxcosx(sinx-cosx)-(1+2sinx))/(sinx-cos^2x)^2#
