If  y=x(logx)log(logx), then  dydxis

y=x(logx)log(logx)⇒log y=(logx)(logx)log(logx)→(1)Taking log of both sides, we get ⇒log (logy)=log(logx)+log(logx)log(logx)Diff. w.r.t x, we get 1logy.1ydydx=1xlogx+2log(logx)logx1x=2log(logx)+1xlogx⇒dydx=yx.logylogx(2log(logx)+1)Substituting the value of log y from (1), we get yx(logx)log(logx)(2log(logx)+1)