Question

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

Answer 1

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

Explanation:

rewrite `y=arcsin(2x+1)` as `siny=2x+1`

Differentiating implicitly we get;
`cosydy/dx=2`
`:. dy/dx=2/cosy`

Using `sin^2A+cos^2A-=1 => (2x+1)^2+cos^2y=1`
`:. 4x^2+4x+1+cos^2y=1`
`:. cos^2y=-4x^2-4x`
`:. cos^2y=4(-x^2-x)`
`cosy=2sqrt(-x^2-x)`

and so `:. dy/dx=2/cosy => dy/dx=2/(2sqrt(-x^2-x))`
Hence, `dy/dx=1/sqrt(-x^2-x)`