# How do you differentiate arcsin(sqrtx)?

Oct 14, 2016

1/(2sqrt(x(1-x))

#### Explanation:

Let $\textcolor{g r e e n}{g \left(x\right) = \sqrt{x}}$ and $f \left(x\right) = \arcsin x$
Then$\textcolor{b l u e}{f \left(\textcolor{g r e e n}{g \left(x\right)}\right) = \arcsin \sqrt{x}}$

Since the given function is a composite function we should differentiate using chain rule.

$\textcolor{red}{f \left(g \left(x\right)\right) '} = \textcolor{red}{f '} \left(\textcolor{g r e e n}{g \left(x\right)}\right) \cdot \textcolor{red}{g ' \left(x\right)}$

Let us compute $\textcolor{red}{f ' \left(\textcolor{g r e e n}{g \left(x\right)}\right)} \mathmr{and} \textcolor{red}{g ' \left(x\right)}$

$f \left(x\right) = \arcsin x$
$f ' \left(x\right) = \frac{1}{\sqrt{1 - {x}^{2}}}$
color(red)(f'(color(green)(g(x)))=1/(sqrt(1-color(green)(g(x))^2))
$f ' \left(\textcolor{g r e e n}{g \left(x\right)}\right) = \frac{1}{\sqrt{1 - {\textcolor{g r e e n}{\sqrt{x}}}^{2}}}$
$\textcolor{red}{f ' \left(g \left(x\right)\right) = \frac{1}{\sqrt{1 - x}}}$

color(red)(g'(x))=?

$\textcolor{g r e e n}{g \left(x\right) = \sqrt{x}}$
$\textcolor{red}{g ' \left(x\right) = \frac{1}{2 \sqrt{x}}}$

$\textcolor{red}{f \left(g \left(x\right)\right) '} = \textcolor{red}{f ' \left(g \left(x\right)\right)} \cdot \textcolor{red}{g ' \left(x\right)}$
$\textcolor{red}{f \left(g \left(x\right)\right) '} = \frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2 \sqrt{x}}$
$\textcolor{red}{f \left(g \left(x\right)\right) '} = \frac{1}{2 \sqrt{x \left(1 - x\right)}}$

Therefore,
color(blue)((arcsinsqrtx)'=1/(2sqrt(x(1-x)))