The correct options are
C yx(logx)log(logx)(2log(logx)+1)
D ylogyxlogx[2log(logx)+1]
y=x(logx)log(logx)
Taking log on both side,
logy=(logx)(logx)log(logx)⋯(1)
Taking log on both side,
log(logy)=log(logx)+log(logx)log(logx)
Differentiating w.r.t. x,
1logy⋅1y⋅dydx=1xlogx+2log(logx)logx⋅1x⇒1logy⋅1y⋅dydx=2log(logx)+1xlogx⇒dydx=yx⋅logylogx(2log(logx)+1)
Using equation (1),
dydx=yx(logx)log(logx)(2log(logx)+1)