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

Jul 31, 2014

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

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

1.) $y = \arccos x$

2.) $\cos y = \cos \left(\arccos x\right)$

3.) $\cos y = x$

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

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

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

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

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

6.) $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sin} \left(\arccos x\right)$

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 $\frac{d}{\mathrm{dx}} \left[\arcsin x\right]$ and $\frac{d}{\mathrm{dx}} \left[\arctan x\right]$, since they appear quite frequently in advanced differentiation problems.