Question

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

Answer 1

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

Explanation:

Use the rule for differentiating arcsine functions:

`d/dx[arcsin(u)]=(u')/sqrt(1-u^2)`

Application of this when `u=2x+1` gives us

`y'=(d/dx[2x+1])/sqrt(1-(2x+1)^2)=2/sqrt(1-(4x^2+4x+1)`

Continue simplification:

`y'=2/sqrt(-4x(x+1))=2/(2sqrt(-x(x+1)))=1/sqrt(-x(x+1))`