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

1 Answer
Jun 6, 2015

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)}#