# What is the derivative of #arcsin(2x+1)#?

##### 1 Answer

#### Explanation:

You can differentiate this function by using implicit differentiation. Start from

#y = arcsin(2x+1)#

This means that you have

#siny = 2x+1#

Now differentiate both sides with respect to

#d/dx(siny) = d/dx(2x+1)#

#cosy * (dy)/dx = 2#

Isolate

#(dy)/dx = 2/cosy#

Use the trigonometric identity

#color(blue)(cos^2y + sin^2y = 1)#

to write

#cos^2y = 1 - sin^2y#

#sqrt(cos^2y) = sqrt(1-sin^2y)#

#cosy = sqrt(1-sin^2y)#

This means that you have

#(dy)/dx = 2/sqrt(1-sin^2y) = 2/(sqrt(1 - (2x+1)^2)#

#(dy)/dx = 2/(sqrt(color(red)(cancel(color(black)(1))) - 4x^2 - 4x - color(red)(cancel(color(black)(1))))) = 2/sqrt(4x(-x-1))#

#(dy)/dx = color(red)(cancel(color(black)(2)))/(color(red)(cancel(color(black)(2)))sqrt(-x(x+1))) = color(green)(1/sqrt(-x(x+1)))#