Question

What is the derivative of `arcsin(x^2/4)`?

Answer 1

`(2x)/sqrt(16-x^4)` for `-2 < x < 2`

Explanation:

Use the Chain Rule, along with the fact that `d/dx(arcsin(x))=1/sqrt(1-x^2)` to get:

`d/dx(arcsin(x^2/4))=1/sqrt(1-(x^2/4)^2) * d/dx(x^2/4)`

`=1/sqrt(1-x^4/16) * x/2=1/sqrt(1-x^4/16) * (2x)/sqrt(16)`

`=(2x)/sqrt(16-x^4)` for `-2 < x < 2`