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

##### 1 Answer
Nov 14, 2016

The derivative is $f ' \left(x\right) = \frac{2 \arcsin x}{\sqrt{1 - {x}^{2}}}$.

#### Explanation:

Let $y = {u}^{2}$ and $u = \arcsin x$. By the chain rule, you have:

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

Let's start by finding the derivative of $u = \arcsin x$.

$u = \arcsin x \to \sin u = x$

Differentiate implicitly:

$\cos u \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) = 1$

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{\cos} u$

Rearrange the identity ${\sin}^{2} x + {\cos}^{2} x = 1$ for $\cos x$ to get $\cos x = \sqrt{1 - {\sin}^{2} x}$.

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {\sin}^{2} u}}$

Since $\sin u = x$, we have:

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

Now to $\frac{\mathrm{dy}}{\mathrm{du}}$. By the product rule, we have, $\frac{\mathrm{dy}}{\mathrm{du}} = 2 u$.

Call the function $y = {\left(\arcsin x\right)}^{2}$ $f \left(x\right)$.

$f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

$f ' \left(x\right) = 2 u \times \frac{1}{\sqrt{1 - {x}^{2}}}$

$f ' \left(x\right) = \frac{2 \arcsin x}{\sqrt{1 - {x}^{2}}}$

Hopefully this helps!