Question

What is the derivative of this function `y=1/arcsin(2x)`?

Answer 1

Start by differentiating `y = arcsin(2x)`.

`y = arcsin(2x) -> siny = 2x`

Through implicit differentiation, we have:

`cosy(dy/dx) = 2`

`dy/dx = 2/(cosy)`

Apply the identity `cosy = sqrt(1 - sin^2(y))`

`dy/dx = 2/sqrt(1 - sin^2y) = 2/sqrt(1 - 4x^2) ->" since siny = 2x"`

So, now we can differentiate the entire expression using the quotient rule.

`dy/dx = (0 xx arcsin(2x) - 2/sqrt(1 - 4x^2))/(arcsin(2x))^2`

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

Hopefully this helps!