Question

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

Answer 1

`2/sqrt(1 - (2x+1)^2)`

Explanation:

differentiate using the `color(blue)" chain rule "`

`d/dx [f(g(x)) ] = f'(g(x)) . g'(x)`

and the standard derivative `D(sin^-1 x) = 1/(sqrt(1 - x^2)`
`"-----------------------------------------------------------------"`

f(g(x)) `= sin^-1 (2x+1) rArr f'(g(x)) = 1/(sqrt(1-(2x+1)^2)`
and g(x) = 2x + 1 → g'(x) = 2
`"-------------------------------------------------------------------"`
Substitute these values into the derivative

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