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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