# Differentiate root 1+tanx/1-tanx?

Jun 8, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} \frac{\frac{\pi}{4} + x}{2 \sqrt{\tan \left(\frac{\pi}{4} + x\right)}}$

#### Explanation:

Let
$y = \sqrt{\frac{1 + \tan x}{1 - \tan x}}$

$y = \sqrt{\frac{1 + \tan x}{1 - \left(1 \times \tan x\right)}}$

$y = \sqrt{\frac{\tan \left(\frac{\pi}{4}\right) + \tan \left(x\right)}{1 - \tan \left(\frac{\pi}{4}\right) \tan \left(x\right)}}$

$y = \sqrt{\tan \left(\frac{\pi}{4} + x\right)}$

Now we can find out the $\frac{\mathrm{dy}}{\mathrm{dx}}$.
We have to apply chain rule to calculate derivative.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{\tan \left(\frac{\pi}{4} + x\right)}} \times d \left(. \tan \left(\frac{\pi}{4} + x\right)\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} \frac{\frac{\pi}{4} + x}{2 \sqrt{\tan \left(\frac{\pi}{4} + x\right)}} \times d \left(\frac{\pi}{4} + x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} \frac{\frac{\pi}{4} + x}{2 \sqrt{\tan \left(\frac{\pi}{4} + x\right)}}$