# How do you differentiate f(x) = arcsin(2x + 1)?

Jan 2, 2016

The final answer is $\frac{1}{\sqrt{- x \left(x + 1\right)}}$
This is found using the standard result for differentiating arcsine, and the chain rule.

#### Explanation:

Solution

$f \left(x\right) = \arcsin \left(2 x + 1\right)$
let
$y = f \left(x\right)$ (I just find it easier to explain in this notation.)

We know that
$\frac{d}{\mathrm{dx}} \left(\arcsin \left(x\right)\right) = \frac{1}{\sqrt{1 - {x}^{2}}}$ (See below for derivation)
apply chain rule
so
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$
and let
$u = 2 x + 1$
so
$y = \arcsin \left(u\right)$
$\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{\sqrt{1 - {u}^{2}}}$

$u = 2 x + 1$
$\frac{\mathrm{du}}{\mathrm{dx}} = 2$

So
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{1}{\sqrt{1 - {u}^{2}}}\right) \cdot 2$

${u}^{2} = 4 {x}^{2} + 4 x + 1$

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

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

dy/dx=2/(2sqrt(-x(x+1))

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

Standard result for the derivative of arcsine function derivation.
let
$y = \arcsin \left(x\right)$
therefore
$x = \sin y$
differentiate implicitly with respect to x
$1 = \cos y \frac{\mathrm{dy}}{\mathrm{dx}}$
rearrange
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} y$
use fundamental trig identity:
${\left(\sin x\right)}^{2} + {\left(\cos x\right)}^{2} = 1$
so
$\cos y = \sqrt{1 - {\left(\sin y\right)}^{2}}$
but
$x = \sin y$
so
$\cos y = \sqrt{1 - {x}^{2}}$
therefore
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {x}^{2}}}$