# How do you find the derivative of Inverse trig function y= arctan(x - sqrt(1+x^2))?

##### 1 Answer
Jun 9, 2018

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \left(1 + {x}^{2}\right)}$

#### Explanation:

Here,

$y = \arctan \left(x - \sqrt{1 + {x}^{2}}\right)$

We take , $x = \cot \theta , w h e r e , \theta \in \left(0 , \pi\right)$

$\implies \theta = a r c \cot x , x \in \mathbb{R} \mathmr{and} \frac{\theta}{2} \in \left(0. \frac{\pi}{2}\right)$

$\implies y = \arctan \left(\cot \theta - \sqrt{1 + {\cot}^{2} \theta}\right)$

$\implies y = \arctan \left(\cot \theta - \csc \theta\right) \ldots \to a p p l y \left(1\right)$

$\implies y = \arctan \left[- \left(\csc \theta - \cot \theta\right)\right]$

$\implies y = - \arctan \left(\csc \theta - \cot \theta\right) \ldots \to A p p l y \left(6\right)$

$\implies y = - \arctan \left(\frac{1}{\sin} \theta - \cos \frac{\theta}{\sin} \theta\right)$

$\implies y = - \arctan \left(\frac{1 - \cos \theta}{\sin} \theta\right) \ldots \to A p p l y \left(2\right) \mathmr{and} \left(3\right)$

$\implies y = - \arctan \left(\frac{2 {\sin}^{2} \left(\frac{\theta}{2}\right)}{2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}\right)$

$\implies y = - \arctan \left(\sin \frac{\frac{\theta}{2}}{\cos} \left(\frac{\theta}{2}\right)\right)$

$\implies y = - \arctan \left(\tan \left(\frac{\theta}{2}\right)\right) , w h e r e , \frac{\theta}{2} \in \left(0 , \frac{\pi}{2}\right)$

$\implies y = - \frac{\theta}{2.} . . \to A p p l y \left(4\right)$

Subst, back , $\theta = a r c \cot \theta$

$y = - \frac{1}{2} a r c \cot \theta$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2} \left(- \frac{1}{1 + {x}^{2}}\right) \ldots \to A p p l y \left(5\right)$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \left(1 + {x}^{2}\right)}$

Note: (Formulas)

$\left(1\right) 1 + {\cot}^{2} \theta = {\csc}^{2} \theta$

$\left(2\right) 1 - \cos \theta = 2 {\sin}^{2} \left(\frac{\theta}{2}\right)$

$\left(3\right) \sin \theta = 2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)$

$\left(4\right) \arctan \left(\tan x\right) = x$

$\left(5\right) \frac{d}{\mathrm{dx}} \left(a r c \cot x\right) = - \frac{1}{1 + {x}^{2}}$

$\left(6\right) \arctan \left(- x\right) = - \arctan x$