Daisy Hatfield

2023-03-27

What is the derivative of $\mathrm{arcsin}\left[{x}^{\frac{1}{2}}\right]$?

anajusthings5mrf

To find the derivative we will need to use the Chain Rule
$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$
We want to find
$\frac{d}{dx}\left(\mathrm{arcsin}\left({x}^{\frac{1}{2}}\right)\right)$
Following the chain rule we let $u={x}^{\frac{1}{2}}$
Deriving u we get
$\frac{du}{dx}=\frac{1}{2}\cdot {x}^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}$
Then, we substitute u in place of x in the original equation and derive to find $\frac{dy}{du}$
$y=\mathrm{arcsin}\left(u\right)$
$\frac{dy}{du}=\frac{1}{\sqrt{1-{u}^{2}}}$
Next, we substitute these derived values into the chain rule to
find $\frac{dy}{dx}$
$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$
$\frac{dy}{dx}=\frac{1}{\sqrt{1-{u}^{2}}}\cdot \frac{1}{2\sqrt{x}}$
Substitute x back into the equation to get the derivative in terms of x only and simplify
$u={x}^{\frac{1}{2}}$
$\frac{dy}{dx}=\frac{1}{\sqrt{1-{\left({x}^{\frac{1}{2}}\right)}^{2}}}\cdot \frac{1}{2\sqrt{x}}$
$\frac{dy}{dx}=\frac{1}{\sqrt{1-x}}\cdot \frac{1}{2\sqrt{x}}$
$\frac{dy}{dx}=\frac{1}{2\sqrt{x}\cdot \sqrt{1-x}}$
$\frac{dy}{dx}=\frac{1}{2\sqrt{x-{x}^{2}}}$

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