Question

How do you differentiate `f(x) = arcsin(2x + 1)`?

Answer 1

The final answer is `1/sqrt(-x(x+1))`
This is found using the standard result for differentiating arcsine, and the chain rule.

Explanation:

Solution

`f(x)=arcsin(2x+1)`
let
`y=f(x)` (I just find it easier to explain in this notation.)

We know that
`d/dx(arcsin(x))=1/sqrt(1-x^2)` (See below for derivation)
apply chain rule
so
`dy/dx=dy/(du)*(du)/(dx)`
and let
`u=2x+1`
so
`y=arcsin(u)`
`dy/(du)=1/sqrt(1-u^2)`

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

So
`dy/dx=(1/sqrt(1-u^2))*2`

`u^2=4x^2 +4x +1`

`dy/dx=2/sqrt(1-4x^2-4x-1)`

`dy/dx=2/sqrt(-4x(x+1))`

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

`dy/dx=1/sqrt(-x(x+1))`

Standard result for the derivative of arcsine function derivation.
let
`y=arcsin(x)`
therefore
`x=siny`
differentiate implicitly with respect to x
`1=cosydy/dx`
rearrange
`dy/dx=1/cosy`
use fundamental trig identity:
`(sinx)^2 + (cosx)^2 =1`
so
`cosy=sqrt(1-(siny)^2)`
but
`x=siny`
so
`cosy=sqrt(1-x^2)`
therefore
`dy/dx=1/sqrt(1-x^2)`