Question

How do you find the derivative of `y=sin^-1(2x+1)`?

Answer 1

The answer is `y'=(2)/sqrt(1-(2x+1)^2`

Explanation:

#d/dx[y]=[1/sqrt(1-(2x+1)^2]*d/dx[2x+1]] #

`y'=(2)/sqrt(1-(2x+1)^2`