Differentiate `x^(tan^-1x)` with respect to `x` :

Explanation

Given that,
`y=x^(tan^-1x)`
taking log on both sides we get,
`=>log y= log x^(tan^-1x)`
`=>log y=tan^-1x* log x`
differentiate on both sides we get,
`=> d/(dx)log y=d/(dx)tan^-1x* log x`
`=>1/y(dy)/(dx)=d/(dx) (tan^-1x)*logx+tan^-1x *d/(dx) (log x)`
`=>(dy)/(dx)=y[logx/(1+x^2)+tan^-1x/x]`
`=>(dy)/(dx)=x^(tan^-1x)[logx/(1+x^2)+tan^-1x/x]`