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(sin^(-1)x)^(x)+(x)^(cos^(-1)x)...

`(sin^(-1)x)^(x)+(x)^(cos^(-1)x)`

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Let `y=(sin^(-1)x)^(x)+(x)^(cos^(-1)x)`
Put `u=(sin^(-1)x)^(x)and upsilon=(x)^(cos^(-1)x)`
Then y = u + `upsilon`
`therefore (dy)/(dx)=(du)/(dx)+(dupsilon)/(dx)" ... (1)"`
Take u `=(sin^(-1)x)^(x)`
`therefore log u=log(sin^(-1)x)^(x)=xlog(sin^(-1)x)`
Differentiating both sides w.r.t. x, we get,
`(1)/(u).(du)/(dx)=(d)/(dx)[xlog(sin^(-1)x)]`
`=x.(d)/(dx)[log(sin^(-1)x)]+log(sin^(-1)x).(d)/(dx)(x)`
`=x xx(1)/(sin^(-1)x).(d)/(dx)(sin^(-1)x)+log(sin^(-1)x)xx1`
`=x xx(1)/(sin^(-1)x)xx(1)/(sqrt(1-x^(2)))+log(sin^(-1)x)`
`therefore (du)/(dx)=u[(x)/(sqrt(1-x^(2)).sin^(-1)x)+log(sin^(-1)x)]`
`=(sin^(-1)x)^(x)[(x)/(sqrt(1-x^(2)).sin^(-1)x)+log(sin^(-1)x)]" ... (2)"`
Also, `upsilon=(x)^(cos^(-1)x)`
`therefore log upsilon=log(x)^(cos^(-1)x)=(cos^(-1)x)(logx)`
Differentiating both sides w.r.t. x, we get,
`(1)/(upsilon).(dupsilon)/(dx)=(d)/(dx)[(cos^(-1)x)(logx)]`
`=(cos^(-1)x).(d)/(dx)(logx)+(logx).(d)/(dx)(cos^(-1)x)`
`=(cos^(-1)x)xx(1)/(x)+(logx)xx(-1)/(sqrt(1-x^(2)))`
`therefore (dupsilon)/(dx)=upsilon[(cos^(-1)x)/(x)-(logx)/(sqrt(1-x^(2)))]`
`=(x)^(cos^(-1)x)[(cos^(-1)x)/(x)-(logx)/(sqrt(1-x^(2)))]`
From (1), (2) and (3), we get,
`(dy)/(dx)=(sin^(-1)x)^(x)[(x)/(sqrt(1-x^(2)).sin^(-1)x)+log(sin^(-1)x)]+(x)^(cos^(-1)x)[(cos^(-1)x)/(x)-(logx)/(sqrt(1-x^(2)))]`
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