# How do you find the derivative of arcsin(1/sqrt(x^2+1))?

Apr 21, 2017

dy/dx=-1/(x^2+1); if x>0,

=1/(x^2+1); if x<0.

N.B.: $y = a r c \sin \left(\frac{1}{\sqrt{{x}^{2} + 1}}\right) , x \in \mathbb{R}$ is not differentiable at $x = 0.$

#### Explanation:

Let, $y = a r c \sin \left(\frac{1}{\sqrt{{x}^{2} + 1}}\right) , x \in \mathbb{R} .$

Let x=cottheta," so that, "theta in (0,pi), &, theta=arc cotx.

Observe that, the Range of $\cot$ fun. is $\mathbb{R} ,$ so we can take,

$x = \cot \theta .$

We will consider the following $2$ Cases :

Case (1) : $x > 0.$

$x = \cot \theta , \theta \in \left(0 , \pi\right) , \mathmr{and} x > 0 \Rightarrow \theta \in \left(0 , \frac{\pi}{2}\right) .$

Also, $y = a r c \sin \left(\frac{1}{\sqrt{{x}^{2} + 1}}\right) = a r c \sin \left(\frac{1}{\sqrt{{\cot}^{2} \theta + 1}}\right)$

$= a r c \sin \left(\frac{1}{\csc} \theta\right) = a r c \sin \left(\sin \theta\right) .$

$\therefore y = a r c \sin \left(\sin \theta\right) , w h e r e , \theta \in \left(0 , \frac{\pi}{2}\right) \subset \left(- \frac{\pi}{2} , \frac{\pi}{2}\right) .$

:.," by Defn. of "arc sin" fun., "y=theta=arc cotx; if x>0

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} a r c \cot x = - \frac{1}{{x}^{2} + 1} , \mathmr{if} x > 0.$

Case (2) : $x < 0.$

Here, $\because x < 0 , \theta \in \left(0 , \pi\right) - \left(0 , \frac{\pi}{2}\right) = \left(\frac{\pi}{2} , \pi\right) , i . e . ,$

$\frac{\pi}{2} < \theta < \pi \Rightarrow - \frac{\pi}{2} > - \theta > - \pi$

$\Rightarrow \pi - \frac{\pi}{2} > \pi - \theta > \pi - \pi , \mathmr{and} , \left(\pi - \theta\right) \in \left(- \frac{\pi}{2} , 0\right)$

Also, $\sin \left(\pi - \theta\right) = \sin \theta .$

Thus, y=arc sin(sintheta)=arc sin(sin(pi-theta),

where, $\left(\pi - \theta\right) \in \left(- \frac{\pi}{2} , 0\right) \subset \left(- \frac{\pi}{2} , \frac{\pi}{2}\right) .$

Hence, by the Defn. of $a r c \sin \text{ fun., } y = \pi - \theta = \pi - a r c \cot x , \mathmr{if} x < 0.$

$\therefore , \frac{\mathrm{dy}}{\mathrm{dx}} = 0 - \left(- \frac{1}{{x}^{2} + 1}\right) = \frac{1}{{x}^{2} + 1} , \mathmr{if} x < 0.$

Altogether, dy/dx=-1/(x^2+1); if x>0,

=1/(x^2+1); if x<0.

N.B.: From the above discussion, we conclude that,

$y = a r c \sin \left(\frac{1}{\sqrt{{x}^{2} + 1}}\right) , x \in \mathbb{R}$ is not differentiable at $x = 0.$

Enjoy Maths.!