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

Question

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

1 Answer

Impact of this question

2756 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-02T14:31:43

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

Explanation:

Solution

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

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

$u = 2 x + 1$ $u=2x+1$
$\frac{d u}{d x} = 2$ $(du)/(dx)=2$

So
$\frac{d y}{d x} = (\frac{1}{\sqrt{1 - u^{2}}}) \cdot 2$ $dy/dx=(1/sqrt(1-u^2))*2$

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

$\frac{d y}{d x} = \frac{2}{\sqrt{1 - 4 x^{2} - 4 x - 1}}$ $dy/dx=2/sqrt(1-4x^2-4x-1)$

$\frac{d y}{d x} = \frac{2}{\sqrt{- 4 x (x + 1)}}$ $dy/dx=2/sqrt(-4x(x+1))$

$\frac{d y}{d x} = \frac{2}{2 \sqrt{- x (x + 1)}}$ $dy/dx=2/(2sqrt(-x(x+1))$

$\frac{d y}{d x} = \frac{1}{\sqrt{- x (x + 1)}}$ $dy/dx=1/sqrt(-x(x+1))$

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

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

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