How do you find the derivative of ##y=arcsin(2x+1)##?

In these cases you need chain rule .

##dy/(dx)## =##d/(dz## x##dz/dx##===>

It's better to take ##d/(dz## & ## dz/dx## separately .

##d/(dz##= ##d/(d(2x+1)##.##arcsin (2x+1)##.

##d/(dz##=##1/sqrt(1-(2x+1)^2##= ##1/(2sqrt(-x^2-x)## ------(1)

##dz/dx##= ##d/dx## .## (2x+1)## =##d/dx 2x## = ##2## ------(2)

Then multiply (1) &(2),

So finally you get ##dy/(dx)##= ##1/(2sqrt(-x^2-x)####2##

##dy/(dx)##=##1/(sqrt(-x^2-x)## ; where ##y## = ##arcsin(2x+1)##