# How do you find the derivative of (arcsin(x))^(2)?

Mar 10, 2018

$\frac{2 \arcsin x}{\sqrt{1 - {x}^{2}}}$

#### Explanation:

We have $\frac{d}{\mathrm{dx}} {\arcsin}^{2} x$

According to the chain rule, $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$, where $u$ is a function within $f$.

Here, $f = {u}^{2}$ where $u = \arcsin x$

So we have $\frac{d}{\mathrm{du}} {u}^{2} \cdot \frac{d}{\mathrm{dx}} \arcsin x$

$2 u \cdot \frac{d}{\mathrm{dx}} \arcsin x$

To find the derivative of $\arcsin x$, take $\arcsin x = {\sin}^{-} 1 x$. We have:

$y = {\sin}^{-} 1 x$, and so:

$\sin y = x$

Taking the derivative of both sides, we have:

$\left(\cos y\right) \cdot y ' = 1$

$y ' = \frac{1}{\cos} y$

We need to write $\cos y$ in terms of $\sin y$, in order to change the variable to $x$.

Remember that ${\sin}^{2} y + {\cos}^{2} y = 1$. So:

${\cos}^{2} y = 1 - {\sin}^{2} y$

$\cos y = \sqrt{1 - {\sin}^{2} y}$

We can take the positive square root, as ${\sin}^{-} 1 x$ is normally given on the interval $- \frac{\pi}{2} , \frac{\pi}{2}$ and as the cosine function is positive in that interval.

Inputting that into our earlier function:

$y ' = \frac{1}{\sqrt{1 - {\sin}^{2} y}}$

$y ' = \frac{1}{\sqrt{1 - {x}^{2}}}$

So as $\frac{d}{\mathrm{dx}} \arcsin x = \frac{1}{\sqrt{1 - {x}^{2}}}$, we can write the derivative of ${\arcsin}^{2} x$ as:

$2 u \cdot \frac{1}{\sqrt{1 - {x}^{2}}}$

But remember that $u = \arcsin x$, so we have:

$\frac{2 \arcsin x}{\sqrt{1 - {x}^{2}}}$

Done.