Question

How do you find the derivative of `arcsin(2x^2)`?

Answer 1

`(4x)/sqrt(1-4x^4`

Explanation:

Here,
`y=sin^(-1)(2x^2)`, take , `u=2x^2`
`y=sin^(-1)u`
`(dy)/(du)=1/sqrt(1-u^2)and(du)/(dx)=4x`
`color(red)((dy)/(dx)=(dy)/(du)*(du)/(dx))=1/sqrt(1-u^2)*4x`
`=>(dy)/(dx)=1/(sqrt(1-(2x^2)^2))*4x=(4x)/sqrt(1-4x^4`