The derivative is 1/{\sqrt{2x}\sqrt(1-2x)}1√2x√1−2x
Let y=arcsin\sqrt{2x}y=arcsin√2x
The goal is to solve for {dy}/{dx}dydx. Take sinsin of each side of the bove equation and get,
sin(y)=\sqrt{2x}sin(y)=√2x
Take the derivative of each side with respect to xx
cos(y){dy}/{dx}=1/\sqrt{2x}cos(y)dydx=1√2x
{dy}/{dx}=1/{\sqrt{2x}cos(y)}dydx=1√2xcos(y)
We need to know what cos(y)cos(y) is. Use sin(y)=\sqrt{2x}/1=\text{opposite}/\text{hypotenuse}sin(y)=√2x1=oppositehypotenuse to get the triangle below.
From the diagram cos(y)={\sqrt{1-2x}}/1cos(y)=√1−2x1. Substitute this into the {dy}/{dx}dydx expression to get,
{dy}/{dx}=1/{\sqrt{2x}\sqrt(1-2x)}dydx=1√2x√1−2x
Alternatively, use the chain rule.
Given,
d/{dz}[arcsin(z)]=1/\sqrt{1-z^2}ddz[arcsin(z)]=1√1−z2
Let z=\sqrt{2x}z=√2x
d/dx[arcsin(\sqrt{2x})]=d/dz[arcsin(z)]dz/dxddx[arcsin(√2x)]=ddz[arcsin(z)]dzdx
Substitute z=\sqrt{2x}z=√2x to get the same answer as before
d/dx[arcsin(\sqrt{2x})]=1/{\sqrt{2x}\sqrt(1-2x)}ddx[arcsin(√2x)]=1√2x√1−2x