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

Explanation

Let `y=u+v,`
where `u=(sinx)^(x) and v=sin^(-1)sqrtx`
Now, `u=(sinx)^(x)`
` rArr logu=x log sin x`
`rArr (1)/(u).(du)/(dx)=x.(cosx)/(sinx)+log sin x.1`
`rArr (du)/(dx)=(sinx)^(x).{x cot +log sin x}.`
And, `v=sin^(-1)sqrtx`
`rArr(dv)/(dx)=(1)/(sqrt(1-x)).(1)/(2)x^(-1//2)`
`=(1)/(2sqrt(x-x^(2)))`
`therefore(dy)/(dx)=(du)/(dx)+(dv)/(dx).`
`(dy)/(dx)=(sinx)^(x).{x cot +log sin x}+(1)/(2sqrt(x-x^(2)))`