What is the derivative of y = arcsin ( x ) ?

This identity can be proven easily by applying sin to both sides of the original equation:
1.) ${\displaystyle y=\arcsin x}$
2.) ${\displaystyle \sin y=\sin \left(\arcsin x\right)}$
3.) ${\displaystyle \sin y=x}$
We continue by using implicit differentiation, keeping in mind to use the chain rule on sin y:
4.) ${\displaystyle \cos y\frac{dy}{dx}=1}$
Solve for ${\displaystyle \frac{dy}{dx}}$:
5.) ${\displaystyle \frac{dy}{dx}=\frac{1}{\cos y}}$
Then, substitution with our original equation yields ${\displaystyle \frac{dy}{dx}}$ in terms of x:
6.) ${\displaystyle \frac{dy}{dx}=\frac{1}{\cos \left(\arcsin x\right)}}$
This may not appear to be a good idea at first, but it can be simplified if the identity is remembered.
${\displaystyle \sin \left(\arccos x\right)=\cos \left(\arcsin x\right)=\sqrt{1-x^2}}$.
7.) ${\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}}$
This is a good definition to memorize, along with ${\displaystyle \frac{d}{dx}\left[\arccos x\right]}$ and ${\displaystyle \frac{d}{dx}\left[\arctan x\right]}$, since they appear quite frequently in differentiation problems.