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)((In x^(x-1))"+2 In x In(In x))"`
B. `(y)/(x)(log x)^(log (log x))(2 log (log x )+1)`
C. `(y)/(" x In x")[(In x )^(2)+2In (In x)]`
D. `(y)/(x)(log y)/(log x)[2 log (log x)+1]`

Answer

B

Explanation

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