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

Mar 1, 2016

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

#### Explanation:

Use chain rule to find the derivative. The derivative of arcsin x is 1/square root of 1-x^2 and then multiply by the derivative of 2x.

Mar 1, 2016

$\setminus \frac{d}{\mathrm{dx}} \left(\setminus \arcsin \left(2 x\right)\right) = \setminus \frac{2}{\setminus \sqrt{1 - 4 {x}^{2}}}$

#### Explanation:

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

Applying chain rule as: $\setminus \frac{\mathrm{df} \left(u\right)}{\mathrm{dx}} = \setminus \frac{\mathrm{df}}{\mathrm{du}} \setminus \cdot \setminus \frac{\mathrm{du}}{\mathrm{dx}}$

Let $2 x = u$

$= \setminus \frac{d}{\mathrm{du}} \left(\setminus \arcsin \left(u\right)\right) . \frac{d}{\mathrm{dx}} \left(2 x\right)$

$\setminus \frac{d}{\mathrm{du}} \left(\setminus \arcsin \left(u\right)\right)$ = $\setminus \frac{1}{\setminus \sqrt{1 - {u}^{2}}}$
{Applying the common derivative : $\setminus \frac{d}{\mathrm{du}} \left(\setminus \arcsin \left(u\right)\right) = \setminus \frac{1}{\setminus \sqrt{1 - {u}^{2}}}$ }

And,
$\setminus \frac{d}{\mathrm{dx}} \left(2 x\right) = 2$

Substituting back $u = 2 x$,

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

Simplifying,
$= \setminus \frac{2}{\setminus \sqrt{1 - 4 {x}^{2}}}$