How do you find the derivative of arcsin((2x)/(1+x^2))?

1 Answer
Jan 24, 2017

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

Explanation:

y=sin^-1((2x)/(x^2+1))

siny=(2x)/(1+x^2)

dy/dx=1/(dx/dy)

dx/dy=cosy

dy/dx=1/cosy

sin^2y+cos^2y=1

cos^2y=1-sin^2y

cosy=sqrt(1-sin^2y)=sqrt(1-(4x^2)/(x^2+1)^2

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