# What is the derivative of arcsin(2x)?

Jun 7, 2015

The derivative of this type of trigonometric function is given by the general rule that follows:

If $y = \arcsin \left(u\right)$, then $y ' = \frac{u '}{\sqrt{1 - {u}^{2}}}$

As in this case our $u = 2 x$, then $u ' = 2$ and we can proceed :)

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

Jun 7, 2015

The derivative of $\arcsin \left(u\right)$ is $\frac{1}{1 + {u}^{2}}$ (See note)

If $u = 2 x$
by the chain rule
$\textcolor{w h i t e}{\text{XXXX}}$$\frac{d \arcsin \left(2 x\right)}{\mathrm{dx}}$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$= \frac{d \arcsin \left(u\right)}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$= \frac{1}{1 + {x}^{2}} \cdot 2$

$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$= \frac{2}{1 + {x}^{2}}$

Note:
$\textcolor{w h i t e}{\text{XXXX}}$I think $\frac{d \arcsin \left(u\right)}{\mathrm{du}} = \frac{1}{1 + {u}^{2}}$
$\textcolor{w h i t e}{\text{XXXX}}$may only be valid if the argument of arcsin is in radians. (Can anyone verify or deny this?)