Question

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

Answer 1

`y=arcsin(2x+1)\\\\siny=2x+1\ (*)\\\\2x=siny-1\\\\x=(1)/(2)siny-(1)/(2)\\\\(dx)/(dy)=(1)/(2)cosy\Rightarrow(dy)/(dx)=(2)/(cosy)=(2)/(√(1-sin^2y))\\\\substitute\ (*)\\\\=(2)/(√(1-(2x+1)^2))=(2)/(√(1-4x^2-4x-1))=(2)/(√(-4x^2-4x))=(2)/(√(4(-x^2-x)))\\\\=(2)/(\sqrt4\cdot√(-x^2-x))=(2)/(2√(-x(x+1)))\\\center\boxed{=(1)/(√(-x(x+1)))}`

Answer 2

Answer is in the attachment below.