#y=tanx^(cot(x/2))#.
Taking natural log, #lny=ln{tanx^(cot(x/2))}, i.e.,#
#lny=cot(x/2)lntanx#.
Diff.ing w.r.t. #x#,
#d/dxlny=d/dx{cot(x/2)lntanx}......(star)#.
Here, by the Chain Rule,
#d/dxlny=d/dylny*dy/dx=1/y*dy/dx...................(star^1)#.
Next, using the Product Rule, we have,
# d/dx{cot(x/2)lntanx}#,
#=cot(x/2)*d/dxlntanx+lntanx*d/dxcot(x/2)#,
#=cot(x/2)*1/tanx*d/dxtanx+(lntanx){-csc^2(x/2)}*d/dx(x/2)#,
#=cot(x/2)*1/tanx*sec^2x-1/2*csc^2(x/2)*lntanx#,
#=cot(x/2)*(1+tan^2x)/tanx-1/2csc^2(x/2)lntanx#,
#=cot(x/2)(1/tanx+tan^2x/tanx)-1/2csc^2(x/2)lntanx#,
#=cot(x/2)(cotx+tanx)-1/2csc^2(x/2)lntanx.............(star^2)#.
Utilising #(star^1) and (star^2)" in "(star)#, we have,
#dy/dx=y{cot(x/2)(cotx+tanx)-1/2csc^2(x/2)lntanx}, or, #
#dy/dx=tanx^(cot(x/2)){cot(x/2)(cotx+tanx)-1/2csc^2(x/2)lntanx}#.
