# How do you find the derivative of arcsin(sqrt(1-x^2)) ?

Nov 29, 2016

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

#### Explanation:

Let $y = \arcsin \left(\sqrt{1 - {x}^{2}}\right)$
Then $\sin y = \sqrt{1 - {x}^{2}} = {\left(1 - {x}^{2}\right)}^{\frac{1}{2}}$

Differentiating Implicitly and applying the chain rule;
$\cos y \frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{1}{2}\right) {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}} \left(- 2 x\right)$
$\therefore \cos y \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{\sqrt{1 - {x}^{2}}} \ldots \left[1\right]$

Then using the fundamental trig identity ${\sin}^{2} A + {\cos}^{2} A \equiv 1$ we have:

${\cos}^{2} y = 1 - {\sin}^{2} y$
${\cos}^{2} y = 1 - {\left({\left(1 - {x}^{2}\right)}^{\frac{1}{2}}\right)}^{2}$
 :. cos^2y=1 - (1-x^2))
$\therefore {\cos}^{2} y = {x}^{2}$
$\therefore \cos y = \pm x$
$\therefore \cos y = | x |$

Substituting into [1] we get:
$\therefore | x | \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{\sqrt{1 - {x}^{2}}}$

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