If y=x(logx)log(logx), then dydxis
yx(1n xlogx−1)+2 1n x 1n (1nx)
yx(logx)log(logx)(2log(logx)+1)
yx 1nx[(1nx)2+21n(1nx)]
yxlogylogx[2log(logx)+1]
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