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

1 Answer
Jan 2, 2016

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

Explanation:

Solution

f(x)=arcsin(2x+1)f(x)=arcsin(2x+1)
let
y=f(x)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)ddx(arcsin(x))=11x2 (See below for derivation)
apply chain rule
so
dy/dx=dy/(du)*(du)/(dx)dydx=dydududx
and let
u=2x+1u=2x+1
so
y=arcsin(u)y=arcsin(u)
dy/(du)=1/sqrt(1-u^2)dydu=11u2

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

So
dy/dx=(1/sqrt(1-u^2))*2dydx=(11u2)2

u^2=4x^2 +4x +1u2=4x2+4x+1

dy/dx=2/sqrt(1-4x^2-4x-1)dydx=214x24x1

dy/dx=2/sqrt(-4x(x+1))dydx=24x(x+1)

dy/dx=2/(2sqrt(-x(x+1))dydx=22x(x+1)

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

Standard result for the derivative of arcsine function derivation.
let
y=arcsin(x)y=arcsin(x)
therefore
x=sinyx=siny
differentiate implicitly with respect to x
1=cosydy/dx1=cosydydx
rearrange
dy/dx=1/cosydydx=1cosy
use fundamental trig identity:
(sinx)^2 + (cosx)^2 =1(sinx)2+(cosx)2=1
so
cosy=sqrt(1-(siny)^2)cosy=1(siny)2
but
x=sinyx=siny
so
cosy=sqrt(1-x^2)cosy=1x2
therefore
dy/dx=1/sqrt(1-x^2)dydx=11x2