Let `y=u+v`, where `y=x^(tanx) and v=sqrt((x^(2)+1)/(x)).`
Now, `u=x^(tanx)`
`rArr logu=(tanx)(logx)`
`rArr(1)/(u).(du)/(dx)=(tanx).(d)/(dx)(logx)+(logx).(d)/(dx)(tanx)`
`" [differentiating w.r.t. x]"`
`=(tanx).(1)/(x)+(logx)sec^(2)x`
`rArr(du)/(dx)=u.[(tanx)/(x)+(logx)sec^(2)x]`
`rArr(du)/(dx)=x^(tanx).{(tanx)/(x)+(logx)sec^(2)x}." ...(i)"`
And, `v=sqrt((x^(2)+1)/(x))`
`rArr log v=(1)/(2).{log (x^(2)+1)-logx}`
`rArr(1)/(v).(dv)/(dx)=(1)/(2).{(2x)/((x^(2)+1))-(1)/(x)}" [differentiating w.r.t. x]"`
`rArr(dv)/(dx)=(v)/(2).{(2x^(2)-(x^(2)+1))/(x(x^(2)+1))}`
`rArr(dv)/(dx)=(1)/(2)sqrt((x^(2)+1)/(x)).{(x^(2)-1)/(x(x^(2)+1))}." ...(ii)"`
`therefore y=u+v`
`rArr(dy)/(dx)=(du)/(dx)+(dv)/(dx)`
`rArr(dy)/(dx)=x^(tanx).{(tanx)/(x)+(logx)sec^(2)x}+(1)/(2).sqrt((x^(2)+1)/(x)).{((x^(2)-1))/(x(x^(2)+1))}.`
