What is the derivative of #arcsin(2x)#?

The derivative of this type of trigonometric function is given by the general rule that follows:

If #y=arcsin(u)#, poi #y'=(u')/(sqrt(1-u^2))#

As in this case our #u=2x#, poi #u'=2# and we can proceed 🙂

#(dy)/(dx)=2/sqrt(1-(2x)^2)=1/sqrt(1-4x^2)#