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

Aug 3, 2014

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {x}^{2}}}$

This identity can be proven easily by applying $\sin$ to both sides of the original equation:

1.) $y = \arcsin x$

2.) $\sin y = \sin \left(\arcsin x\right)$

3.) $\sin y = x$

We continue by using implicit differentiation, keeping in mind to use the chain rule on $\sin y$:

4.) $\cos y \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

5.) $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} y$

Now, substitution with our original equation yields $\frac{\mathrm{dy}}{\mathrm{dx}}$ in terms of $x$:

6.) $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} \left(\arcsin x\right)$

At first this might not look all that great, but it can be simplified if one recalls the identity
$\sin \left(\arccos x\right) = \cos \left(\arcsin x\right) = \sqrt{1 - {x}^{2}}$.

7.) $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {x}^{2}}}$

This is a good definition to memorize, along with $\frac{d}{\mathrm{dx}} \left[\arccos x\right]$ and $\frac{d}{\mathrm{dx}} \left[\arctan x\right]$, since they appear quite frequently in differentiation problems.