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

1 Answer
Max
Jan 2, 2016

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)

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