The given function is \( \displaystyle y = \sqrt{\frac{1 – \tan x}{1 + \tan x}} \) and we need to find the derivative of this function with respect to \( x \).
Solution:
We know that: \( \displaystyle \tan(a-b) = \frac{\tan a – \tan b}{1+ \tan a \cdot \tan b} \)
Therefore, \( \displaystyle \tan\left(\frac{\pi}{4} – x\right) = \frac{\tan \frac{\pi}{4} – \tan x}{1 + \tan \frac{\pi}{4} \cdot \tan x} = \frac{1 – \tan x}{1 + \tan x}, \because \tan \frac{\pi}{4} = 1 \)
Hence, \( \displaystyle y = \sqrt{\frac{1 – \tan x}{1 + \tan x}} = \sqrt{\tan\left(\frac{\pi}{4} – x\right)} \)
Let \( \displaystyle u = \tan\left(\frac{\pi}{4} – x\right) \) and \( \displaystyle v = \frac{\pi}{4} – x \).
Using the chain rule: \( \displaystyle \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx} \dots (1)\).
Compute \(\displaystyle \frac{dy}{du} \):
\( \begin{aligned} &\frac{dy}{du} = \frac{d (\sqrt{u})}{du} = \frac{1}{2}u^\frac{-1}{2} \\ & = \frac{1}{2} \left( \tan \left(\frac{\pi}{4} – x\right) \right)^{\frac{-1}{2}} \\ &= \frac{1}{2} \left( \frac{1 – \tan x}{1 + \tan x} \right)^{\frac{-1}{2}} \\ &= \frac{1}{2} \sqrt{\frac{1 + \tan x}{1 – \tan x}} \\ \end{aligned} \)Compute \(\displaystyle \frac{du}{dv} \):
\( \begin{aligned} \frac{du}{dv} = \frac{d}{dv} \tan v = \sec^{2}v = \sec^{2} \left(\frac{\pi}{4} – x \right) \\ \end{aligned} \)Compute \(\displaystyle \frac{dv}{dx} \):
\( \begin{aligned} \frac{dv}{dx} = \frac{d}{dx} \left( \frac{\pi}{4} – x \right) = -1 \\ \end{aligned} \)Put all values in (1):
\( \begin{aligned} &\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{1 + \tan x}{1 – \tan x}} \cdot \sec^{2} \left(\frac{\pi}{4} – x \right) \cdot -1 \\ &= – \frac{1}{2} \sqrt{\frac{1 + \tan x}{1 – \tan x}} \cdot \sec^{2} \left(\frac{\pi}{4} – x \right) \\ \end{aligned} \)Thus, the derivative of the function is:
\( \boxed{\frac{dy}{dx} = – \frac{1}{2} \sqrt{\frac{1 + \tan x}{1 – \tan x}} \cdot \sec^{2} \left(\frac{\pi}{4} – x \right) } \\ \)Please let me know in the comments if you find any errors in this solution.