If `y=x^(logx)+(logx)^x` then find `(dy)/(dx)`

Explanation

Given that,
`y=x^(logx)+(logx)^x`
Let `y=u+v`, where `u=(logx)^(x) and v=x^(logx)`
Then, `(dy)/(dx)=(du)/(dx)+(dv)/(dx).`
Now, `u=(logx)^(x) rArr logu=xlog(logx)`
`rArr (1)/(u).(du)/(dx)=x.(1)/(logx).(1)/(x)+log(logx).1`
`rArr (du)/(dx)=u.{(1)/(logx)+log(logx)}`
`rArr(du)/(dx)=(logx)^(x).{(1)/(logx)+log(logx)}.`
And, `v=x^(logx)rArr log v=(logx)(logx)=(logx)^(2)`
`rArr (1)/(v).(dv)/(dx)=(2(logx))/(x)`
`rArr (dv)/(dx)=v.(2(logx))/(x)=e^((logx)^(2))*(2(logx))/(x)`