Question

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

Answer 1

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

Explanation:

Let `y = (arcsinx)^2 => y^(1/2) = arcsinx`
`:. sin(y^(1/2)) = x`

Differentiating wrt `x`;
`:. cos(y^(1/2))(1/2y^(-1/2))dy/dx = 1`

`:. dy/dx = 2/(cos(y^(1/2))(y^(-1/2)))`

`:. dy/dx = (2y^(1/2))/(cos(y^(1/2)))`

Now `sin^2A + cos^2A -= 1 => cosA = sqrt(1 - sin^2A`
So, `cos(y^(1/2)) = sqrt(1 - sin^2(y^(1/2))`
`:. cos(y^(1/2)) = sqrt(1 - x^2)`

Hence, `dy/dx = (2arcsinx)/sqrt(1 - x^2)`