# What is the derivative of  ArcSin[x^(1/2)]?

Dec 4, 2016

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

#### Explanation:

When tackling the derivative of inverse trig functions. I prefer to rearrange and use Implicit differentiation as I always get the inverse derivatives muddled up, and this way I do not need to remember the inverse derivatives. If you can remember the inverse derivatives then you use the chain rule.

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

Differentiate Implicitly:

$\cos y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {x}^{- \frac{1}{2}}$
$\therefore \cos y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{x}}$ ..... [1]

Using the $\sin \text{/} \cos$ identity;

${\sin}^{2} y + {\cos}^{2} y \equiv 1$
$\therefore {\cos}^{2} y = 1 - {\sin}^{2} y$
$\therefore {\cos}^{2} y = 1 - {\left(\sqrt{x}\right)}^{2}$
$\therefore {\cos}^{2} y = 1 - x$
$\therefore \cos y = \sqrt{1 - x}$

Substituting into [1]
$\therefore \sqrt{1 - x} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{x}}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{x} \sqrt{1 - x}}$