If `y=x^(logx)^(("log"(logdotx))),t h e n(dy)/(dx)i s` `y/x(1n x^(oox-1))+21 nx1n(1nx))` `y/x(logx)^("log"(logx))(2log(logx)+1)` `y/(x1nx)[(1nx)^2+21 n(1nx)]` `y/x(logy)/(logx)[2log(logx)+1]`

Options

A. `(y)/(x)(lnx^(logx-1))+2lnxln(lnx)`
B. `(y)/(x)(logx)^(log(logx))(2log(logx)+1)`
C. `[(lnx)^(2)+2ln(lnx)]`
D. `(y)/(x)(logy)/(logx)(2log(logx)+1)`

Answer

B, D

Explanation

`y=x^((logx)^(log(logx)))`
`rArr" "logy=(logx)(logx)^(log(logx))" (1)"`
Taking log on both sides, we get
`log(logy)=log(logx)+log(logx)log(logx)`
Differentiating w.r.t., we get
`(1)/(logy).(1)/(y)(dy)/(dx)=(1)/(xlogx)+(2log(logx))/(logx)(1)/(x)`
`=(2log(logx)+1)/(xlogx)`
`rArr" "(dy)/(dx)=(y)/(x).(logy)/(logx)(2log(logx)+1)`
Substituting the value of log y from (1), we get
`(dy)/(dx)=(y)/(x)(logx)^(log(logx))(2log(logx)+1)`