# How do you find the derivative of Cos[arcsin(x)] ?

Jun 8, 2017

$\frac{d}{\mathrm{dx}} \cos \left(\arcsin \left(x\right)\right) = - \frac{x}{\sqrt{1 - {x}^{2}}}$

#### Explanation:

Note that:

${\cos}^{2} \theta + {\sin}^{2} \theta = 1$

So for any $\theta$, we have:

$\cos \left(\theta\right) = \pm \sqrt{1 - \sin \left(\theta\right)}$

Note that for any real $x \in \left[- 1 , 1\right]$, by definition:

$\arcsin \left(x\right) \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

Further note that if $\theta \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$ then:

$\cos \theta \ge 0$

Hence:

$\cos \left(\arcsin \left(x\right)\right) = \sqrt{1 - {x}^{2}}$

for any $x \in \left[- 1 , 1\right]$

Outside this interval $\cos \left(\arcsin \left(x\right)\right)$ takes non-real complex values.

So if we are dealing with this purely as a real valued function, then this is the only definition we need.

Then:

$\frac{d}{\mathrm{dx}} \cos \left(\arcsin \left(x\right)\right) = \frac{d}{\mathrm{dx}} {\left(1 - {x}^{2}\right)}^{\frac{1}{2}}$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \cos \left(\arcsin \left(x\right)\right)} = \frac{1}{2} {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}} \cdot \left(- 2 x\right)$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \cos \left(\arcsin \left(x\right)\right)} = - \frac{x}{\sqrt{1 - {x}^{2}}}$