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

Jun 6, 2015

To find the derivative we will need to use the Chain Rule

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

We want to find

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

Following the chain rule we let $u = {x}^{\frac{1}{2}}$

Deriving u we get

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

Now we substitute u in place of x in the original equation and derive to find $\frac{\mathrm{dy}}{\mathrm{du}}$

$y = \arcsin \left(u\right)$

(dy)/(du)=1/(sqrt(1-u^2)

Now we substitute these derived values into the chain rule to
find $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

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

Substitute x back into the equation to get the derivative in terms of x only and simplify

$u = {x}^{\frac{1}{2}}$

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

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

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

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