Question

(d/dx)[(xcosa)^2+2x(sinb)^-2]=

Answer 1

Dependin on whether you assume that a and b are functions of x, or constants, you get two very different answers. First we'll approach this with the assumption that both a and b are functions of x.
----------------------------------------------------------
d/dx = 2(xcos(a)*([-xsin(a)*da/dx+cos(a)]+[2sin(b)-2xcos(b)*db/dx])/sin^2(b) ----------------
----------------------------------------
If you assume that both of the trig ratios are constants, then it ends up as d/dx= 2(xcos(a))*cos(a) +2/sin(b)