# How do you differentiate Arctan ((x-1)/(x+1))?

##### 2 Answers
Apr 14, 2017

$\frac{1}{1 + {x}^{2}} , x \ne - 1 , x \in \mathbb{R} .$

#### Explanation:

Let, $y = a r c \tan \left\{\frac{x - 1}{x + 1}\right\}$

Clearly, $x \in \mathbb{R} - \left\{- 1\right\} .$

Recall that the Range of $\tan$ fun. is $\mathbb{R} ,$ so, we can, suppose,

$x = \tan \theta , \text{ so that, } \theta = a r c \tan x .$

$\text{Also, } x \in \mathbb{R} - \left\{- 1\right\} \Rightarrow \theta \in \mathbb{R} - \left\{n \pi - \frac{\pi}{4} : n \in \mathbb{Z}\right\} .$

Now, $\tan \left\{\frac{x - 1}{x + 1}\right\} = \frac{\tan \theta - \tan \left(\frac{\pi}{4}\right)}{1 + \tan \theta \cdot \tan \left(\frac{\pi}{4}\right)} ,$

$= \tan \left(\theta - \frac{\pi}{4}\right)$

$\Rightarrow y = a r c \tan \left\{\frac{x - 1}{x + 1}\right\} = a r c \tan \left\{\tan \left(\theta - \frac{\pi}{4}\right)\right\} , i . e . ,$

$y = \theta - \frac{\pi}{4} = a r c \tan x - \frac{\pi}{4} , x \in \mathbb{R} - \left\{- 1\right\} .$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + {x}^{2}} - 0 = \frac{1}{1 + {x}^{2}} , x \in \mathbb{R} - \left\{- 1\right\} .$

Enjoy Maths.!

Apr 16, 2017

Recall that the derivative of $\arctan \left(x\right)$ is $\frac{1}{1 + {x}^{2}}$. Then, by the chain rule, the derivative of $\arctan \left(f \left(x\right)\right)$ is $\frac{1}{1 + f {\left(x\right)}^{2}} \cdot f ' \left(x\right)$.

Here, this shows us that:

$\frac{d}{\mathrm{dx}} \arctan \left(\frac{x - 1}{x + 1}\right) = \frac{1}{1 + {\left(\frac{x - 1}{x + 1}\right)}^{2}} \frac{d}{\mathrm{dx}} \left(\frac{x - 1}{x + 1}\right)$

First simplifying the fraction:

$= {\left(x + 1\right)}^{2} / \left({\left(x + 1\right)}^{2} + {\left(x - 1\right)}^{2}\right) \frac{d}{\mathrm{dx}} \left(\frac{x - 1}{x + 1}\right)$

Using the quotient rule:

$= {\left(x + 1\right)}^{2} / \left({\left(x + 1\right)}^{2} + {\left(x - 1\right)}^{2}\right) \left(\frac{\left(x + 1\right) - \left(x - 1\right)}{x + 1} ^ 2\right)$

$= \frac{2}{{\left(x + 1\right)}^{2} + {\left(x - 1\right)}^{2}}$

$= \frac{2}{\left({x}^{2} + 2 x + 1\right) + \left({x}^{2} - 2 x + 1\right)}$

$= \frac{2}{2 {x}^{2} + 2}$

$= \frac{1}{{x}^{2} + 1}$