What is the derivative of arcsin sqrt(2x)arcsin2x?

1 Answer
Jun 6, 2015

The derivative is 1/{\sqrt{2x}\sqrt(1-2x)}12x12x

Let y=arcsin\sqrt{2x}y=arcsin2x

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=12x

{dy}/{dx}=1/{\sqrt{2x}cos(y)}dydx=12xcos(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.enter image source here

From the diagram cos(y)={\sqrt{1-2x}}/1cos(y)=12x1. Substitute this into the {dy}/{dx}dydx expression to get,

{dy}/{dx}=1/{\sqrt{2x}\sqrt(1-2x)}dydx=12x12x

Alternatively, use the chain rule.

Given,

d/{dz}[arcsin(z)]=1/\sqrt{1-z^2}ddz[arcsin(z)]=11z2

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)]=12x12x