Question

What is the derivative of `arcsin(2x+1)`?

Answer 1

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

Explanation:

You can differentiate this function by using implicit differentiation. Start from

`y = arcsin(2x+1)`

This means that you have

`siny = 2x+1`

Now differentiate both sides with respect to `x`

`d/dx(siny) = d/dx(2x+1)`

`cosy * (dy)/dx = 2`

Isolate `(dy)/dx` on one side of the equation

`(dy)/dx = 2/cosy`

Use the trigonometric identity

`color(blue)(cos^2y + sin^2y = 1)`

to write `cosy` as a function of `siny`

`cos^2y = 1 - sin^2y`

`sqrt(cos^2y) = sqrt(1-sin^2y)`

`cosy = sqrt(1-sin^2y)`

This means that you have

`(dy)/dx = 2/sqrt(1-sin^2y) = 2/(sqrt(1 - (2x+1)^2)`

`(dy)/dx = 2/(sqrt(color(red)(cancel(color(black)(1))) - 4x^2 - 4x - color(red)(cancel(color(black)(1))))) = 2/sqrt(4x(-x-1))`

`(dy)/dx = color(red)(cancel(color(black)(2)))/(color(red)(cancel(color(black)(2)))sqrt(-x(x+1))) = color(green)(1/sqrt(-x(x+1)))`