The derivative is 1/{\sqrt{2x}\sqrt(1-2x)}
Let y=arcsin\sqrt{2x}
The goal is to solve for {dy}/{dx}. Take sin of each side of the bove equation and get,
sin(y)=\sqrt{2x}
Take the derivative of each side with respect to x
cos(y){dy}/{dx}=1/\sqrt{2x}
{dy}/{dx}=1/{\sqrt{2x}cos(y)}
We need to know what cos(y) is. Use sin(y)=\sqrt{2x}/1=\text{opposite}/\text{hypotenuse} to get the triangle below.![enter image source here]()
From the diagram cos(y)={\sqrt{1-2x}}/1. Substitute this into the {dy}/{dx} expression to get,
{dy}/{dx}=1/{\sqrt{2x}\sqrt(1-2x)}
Alternatively, use the chain rule.
Given,
d/{dz}[arcsin(z)]=1/\sqrt{1-z^2}
Let z=\sqrt{2x}
d/dx[arcsin(\sqrt{2x})]=d/dz[arcsin(z)]dz/dx
Substitute z=\sqrt{2x} to get the same answer as before
d/dx[arcsin(\sqrt{2x})]=1/{\sqrt{2x}\sqrt(1-2x)}
