dy  dy = 2{tanx.cosec5x[3cos8x.sec3x.tan3x-8sin8x.sec3x] + cos8x.sec3x[sec² x.cosex5x-5tanx.cosec5x.cot5x] }               dx                         is the answer right? plz lemme know!! simplify it further if needed!!! cheers!!                                                                                                                                                                                                                                                                                 

dx

dy = 2{tanx.cosec5x[3cos8x.sec3x.tan3x-8sin8x.sec3x] + cos8x.sec3x[sec² x.cosex5x-5tanx.cosec5x.cot5x] }               dx                         is the answer right? plz lemme know!! simplify it further if needed!!! cheers!! itz nt getting printed properly.... jus check 2{[.......................]} on wards.....that is dy/dx

If you expand the sin2x in numerator as sin2x = 2sinxcosx, then you eliminate two of the denominator terms. Take cotx in the Numerator as tanx. The expression becomes

cosec5x cos8x sec3x tanx

Differentiate using this rule

d(abce)/dx = bce(da/dx)  + ace(db/dx) + abe(dc/dx) + abc(de/dx)

y = (cosec(5x).sin(2x).cos(8x).sec(3x))/(cos(x).sin(x).(cos(x)/sin(x)))

y = (cosec(5x).sin(2x).cos(8x).sec(3x))/(cos²(x))

y = 2.(cosec(5x).tan(x).cos(8x).sec(3x)) (writing sin(2x) as 2.sin(x).cos(x))

(dy/dx) = 2.5(-cosec(5x).cot(5x))tan(x).cos(8x).sec(3x) +

                2.cosec(5x).sec²(x).cos(8x).sec(3x) +

                2.cosec(5x).tan(x).(-8sin(8x)).sec(3x) +

                2.cosec(5x).tan(x).cos(8x).3sec(x).tan(x)

Formula used:- If y = u.v.w then dy/dx = (du/dx).v.w + u.(dv/dx).w + u.v.(dw/dx)