Question

What is the derivative of `arcsin(2x)`?

Answer 1

The derivative of this type of trigonometric function is given by the general rule that follows:

If `y=arcsin(u)`, then `y'=(u')/(sqrt(1-u^2))`

As in this case our `u=2x`, then `u'=2` and we can proceed :)

`(dy)/(dx)=2/sqrt(1-(2x)^2)=1/sqrt(1-4x^2)`

Answer 2

The derivative of `arcsin(u)` is `1/(1+u^2)` (See note)

If `u = 2x`
by the chain rule
`color(white)("XXXX")` `(d arcsin(2x))/(dx)`
`color(white)("XXXX")` `color(white)("XXXX")` `=(d arcsin(u))/(du) * (du)/(dx)`

`color(white)("XXXX")` `color(white)("XXXX")` `= 1/(1+x^2)*2`

`color(white)("XXXX")` `color(white)("XXXX")` `= 2/(1+x^2)`

Note:
`color(white)("XXXX")`I think `(d arcsin(u))/(du) = 1/(1+u^2)`
`color(white)("XXXX")`may only be valid if the argument of arcsin is in radians. (Can anyone verify or deny this?)