# What is the derivative of  arcsin(x^(1/2))?

Jun 30, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \left(\sqrt{x \left(1 - x\right)}\right)}$

#### Explanation:

First I'm going to walk you through a nice way of deriving the inverse trig derivatives.

Start with $y = \arcsin \left(x\right)$ ,this allows us to construct a triangle shown below.

Rearranging we get that $x = \sin \left(y\right)$

Hence, $\frac{\mathrm{dx}}{\mathrm{dy}} = \cos \left(y\right)$

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

We obtain that:

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

Now that we know the general form, we use the chain rule.

For $y \left(u \left(x\right)\right) , \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

So $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {\left({x}^{\frac{1}{2}}\right)}^{2}}} \cdot \frac{d}{\mathrm{dx}} \left({x}^{\frac{1}{2}}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \left(\sqrt{x \left(1 - x\right)}\right)}$