Find `(dy)/(dx)`, when:
`y=(sinx)^(x)+sin^(-1)sqrtx`

Answer

`(sinx)^(x).{x cot x+log sin x}+(1)/(2sqrt(x-x^(2)))`

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)sqrtxrArr(dv)/(dx)=(1)/(sqrt(1-x)).(1)/(2)x^(-1//2)=(1)/(2sqrt(x-x^(2)))`
`therefore(dy)/(dx)=(du)/(dx)+(dv)/(dx).`