Question

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

Answer 1

The derivative is `f'(x) = (2arcsinx)/sqrt(1 - x^2)`.

Explanation:

Let `y = u^2` and `u = arcsinx`. By the chain rule, you have:

`dy/dx = dy/(du) xx (du)/dx`

Let's start by finding the derivative of `u = arcsinx`.

`u = arcsinx -> sinu = x`

Differentiate implicitly:

`cosu((du)/dx) = 1`

`(du)/dx = 1/cosu`

Rearrange the identity `sin^2x + cos^2x = 1` for `cosx` to get `cosx = sqrt(1 - sin^2x)`.

`(du)/dx= 1/sqrt(1 - sin^2u)`

Since `sinu = x`, we have:

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

Now to `dy/(du)`. By the product rule, we have, `dy/(du) = 2u`.

Call the function `y = (arcsinx)^2` `f(x)`.

`f'(x) = dy/(du) xx (du)/dx`

`f'(x) = 2u xx 1/sqrt(1 - x^2)`

`f'(x) = (2arcsinx)/sqrt(1- x^2)`

Hopefully this helps!