Let u=x2-cos(x), du/dx=2x+sin(x).
Let v=sin(x)+8x, dv/dx=cos(x)+8.
(d/dx)(u/v)=(vdu/dx-udv/dx)/v2=
((sin(x)+8x)(2x+sin(x))-(x2-cos(x))(cos(x)+8))/(sin(x)+8x)2.
So (d/dx)((x2-cos(x))/(sin(x)+8x))=
((sin(x)+8x)(2x+sin(x))-(x2-cos(x))(cos(x)+8))/(sin(x)+8x)2.
Expanding these terms:
(sin2(x)+10xsin(x)+16x2-(8x2+x2cos(x)-8cos(x)-cos2(x)))/(sin(x)+8x)2,
(sin2(x)+cos2(x)+8x2-x2cos(x)+10xsin(x)+8cos(x))/(sin(x)+8x)2,
(1+8x2-x2cos(x)+10xsin(x)+8cos(x))/(sin(x)+8x)2.
So (d/dx)((x2-cos(x))/(sin(x)+8x))=
(1+8x2-x2cos(x)+10xsin(x)+8cos(x))/(sin(x)+8x)2.