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

1 Answer
Jan 2, 2016

The final answer is 1x(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
ddx(arcsin(x))=11x2 (See below for derivation)
apply chain rule
so
dydx=dydududx
and let
u=2x+1
so
y=arcsin(u)
dydu=11u2

u=2x+1
dudx=2

So
dydx=(11u2)2

u2=4x2+4x+1

dydx=214x24x1

dydx=24x(x+1)

dydx=22x(x+1)

dydx=1x(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=cosydydx
rearrange
dydx=1cosy
use fundamental trig identity:
(sinx)2+(cosx)2=1
so
cosy=1(siny)2
but
x=siny
so
cosy=1x2
therefore
dydx=11x2