If `y=(sinx)^(x)+sin^(-1)sqrtx,` find `(dy)/(dx).`

Explanation

Let `(sinx)^(x)=u and sin^(-1)sqrtx=v`. Then,
`y=u+v rArr (dy)/(dx)=(du)/(dx)+(dv)/(dx)." …(i)"`
Now, `u=(sinx)^(x) rArr logu = xlog (sinx)`
`rArr (1)/(u).(du)/(dx)=x.(1)/(sinx).cosx+log(sinx).1`
`rArr (du)/(dx)=u{x cot x+log(sinx)}`
`rArr (du)/(dx)=(sinx)^(x).{x cot x+log(sinx)}." ....(ii)"`
`v=sin^(-1)sqrtxrArr(dv)/(dx)=(1)/(sqrt((1-x))).(1)/(2)x^(-1//2)=(1)/(2sqrtxsqrt((1-x)))=(1)/(2sqrt(x-x^(2)))." ...(iii)"`
Using (ii) and (iii) in (i), we get
`(dy)/(dx)=(sinx)^(x){x cot x+log(sinx)}+(1)/(2sqrt(x-x^(2))).`