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

Question

2022 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-17T19:06:48

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

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

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