Differentiate w.r.t. x the function `(logx)^(logx), x >1`

Explanation

Let `y=(logx)^(logx)`

Taking logarithm on both the sides,we obtain

`log y =log x*log(logx)`

Differentiating both sides with respect to x, we obtain

`1/y``dy/dx`=`d/dx[logx*log(logx)]`
=>`1/y``dy/dx`=`log(logx)*d/dx(logx)+logx*d/dx[log(logx)]`

=>`dy/dx=y[log(logx)*1/x+log*1/logx*d/dx(logx)]`

=>`dy/dx=y[1/xlog(logx)+1/x]`

=>`dy/dx=(logx)^logx``[1/x+log(logx)/x]`