Differentiate the function `(sin x)^x+ sin^1 x` with respect to x.

Explanation

`"Let " y = ("sin"x)^(x) + "sin"^(-1) sqrt(x)`
`"Let " u = ("sin"x)^(x) "and " v = "sin"^(-1) sqrt(x)`
`therefore y = u+v`
`rArr (dy)/(dx) = (du)/(dx) + (dv)/(dx) " " .....(1)`
`"Now", u = ("sin" x)^(x)`
`rArr "log"u = "log"("sin" x)^(x) = x "log " "sin"x`
`rArr (1)/(u) (du)/(dx) = x (d)/(dx) "log sin" x + "log sin"x (d)/(dx)x`
`rArr (du)/(dx) = u[x* ("cos"x)/("sin"x) + "log sin" x * 1]`
`rArr (du)/(dx) = ("sin"x)^(x) [x "cot" x + "log sin"x]`
`"and " v="sin"^(-1) sqrt(x)`
`rArr (dv)/(dx) = (d)/(dx)"sin"^(-1) sqrt(x)`
`= (1)/(sqrt(1-(sqrt(x))^(2))) (d)/(dx)sqrt(x) = (1)/(2sqrt(x)(1-x))`
From equation (1)
`(dy)/(dx) = ("sin" x)^(x)[x "cot"x + "log sin"x] + (1)/(2sqrt(x)(1-x))`