Question

How do you find the derivative of `arcsin(2x + 1)`?

Answer 1

`1/sqrt(-x(x+1))`

Explanation:

Use the formula: `d/dx arcsin (u/a) = 1/sqrt(a^2-u^2) (du)/dx`

From the equation `y = arcsin(2x+1)`:

`a = 1` and `u = 2x+1` and `(du)/dx = 2`

so `y' = 2/sqrt(1^2-(2x+1)^2)`

Simplify the square root function:
`1^2-(2x+1)^2 = 1-(4x^2+4x+1) = -4x^2-4x`
`= -4x(x+1)`

so `y' = 2/sqrt(-4x(x+1)) = 2/(2sqrt(-x(x+1)) )= 1/sqrt(-x(x+1))`