# What is the derivative of f(x)=arctan[(x+1)/(x-1)]?

May 14, 2018

$\Rightarrow f ' \left(x\right) = - \frac{1}{{x}^{2} + 1} , \left(x \ne 1\right)$.

#### Explanation:

Observe that,

$\tan \left(\frac{\pi}{4} + \theta\right) = \frac{\tan \left(\frac{\pi}{4}\right) + \tan \theta}{1 - \tan \left(\frac{\pi}{4}\right) \tan \theta} = \frac{1 + \tan \theta}{1 - \tan \theta}$.

$\therefore \tan \left(- \left(\frac{\pi}{4} + \theta\right)\right) = - \tan \left(\frac{\pi}{4} + \theta\right) = \frac{\tan \theta + 1}{\tan \theta - 1}$.

Recall that, the range of $\tan$ function is $\mathbb{R}$.

So, we can very well take $x = \tan \theta$ and get,

$f \left(x\right) = a r c \tan \left\{\frac{x + 1}{x - 1}\right\} = a r c \tan \left\{\frac{\tan \theta + 1}{\tan \theta - 1}\right\}$,

=arc tan{tan(-(pi/4+theta)},

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

$\Rightarrow f \left(x\right) = - \frac{\pi}{4} - a r c \tan x , \left(x \ne 1\right)$.

$\Rightarrow f ' \left(x\right) = 0 - \frac{1}{{x}^{2} + 1} = - \frac{1}{{x}^{2} + 1} , \left(x \ne 1\right)$, is the

desired derivative!