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

Explanation

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