Find `(dy)/(dx)` for the function: `y=a^(sin^(-1)x)^(2)`

Answer

`(2log a sin^(-1)x)/(sqrt(1-x^(2)))a(sin^(-1)x)^(2)`

Explanation

`"Let "y=a^((sin^(-1)x)^(2))`.
Using chain rule, we get
`(dy)/(dx)=(d)/(dx){a^((sin^(-1)x)^(2))}`
`=a^((sin^(-1)x)^(2))loga(d)/(x){(sin^(-1)x)^(2)}`
`=a^((sin^(-1)x)^(2))(log a) 2 (sin ^(-1)x)^(1)(d)/(dx)(sin^(-1)x)`
`=a^((sin^(-1)x)^(2))(log a)2 sin^(-1)x(1)/(sqrt(1-x^(2)))`
`=(2 log a sin^(-1)x)/(sqrt(1-x^(2)))a^((sin^(-1)x)^(2))`