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

$\frac{d}{d x}\left(\sqrt{\frac{1-\tan x}{1+\tan x}}\right)=$
  1. $\frac{\sec ^2 x}{(1+\tan x)^{3 / 2}(1-\tan x)^{1 / 2}}$
  2. $\frac{-\sec ^2 x}{\left(1-\tan ^2 x\right)^{1 / 2}}$
  3. $\frac{\sec ^2 x}{\left(1-\tan ^2 x\right)^{1 / 2}}$
  4. $\frac{-\sec ^2 x}{(1+\tan x)^{3 / 2}(1-\tan x)^{1 / 2}}$

Solution

$\begin{aligned} & \frac{d}{d x}\left(\sqrt{\frac{1-\tan x}{1+\tan x}}\right)=\frac{1}{\sqrt[2]{\frac{1-\tan x}{1+\tan x}}} \times \frac{(1+\tan x)\left(0-\sec ^2 x\right)-(1-\tan x)\left(0-\sec ^2 x\right)}{(1+\tan x)^2} \\ & =\frac{-2 \sec ^2 x}{2 \sqrt{1-\tan x}(1+\tan x)^{3 / 2}} \\ & =\frac{-\sec ^2 x}{(1-\tan x)^{1 / 2} \cdot(1+\tan x)^{3 / 2}}\end{aligned}$

Asked in: MHT CET 2022 (05 Aug Shift 2)

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