The derivative of
Question:

The derivative of $\tan ^{-1}\left(\frac{\sin x-\cos x}{\sin x+\cos x}\right)$, with respect to $\frac{x}{2}$,

where $\left(x \in\left(0, \frac{\pi}{2}\right)\right)$ is :

1. (1) 1

2. (2) $\frac{2}{3}$

3. (3) $\frac{1}{2}$

4. (4) 2

Correct Option: , 4

Solution:

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

$=-\tan ^{-1}\left(\tan \left(\frac{\pi}{4}-x\right)\right) \quad\left[\because \frac{\pi}{4}-x \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)\right]$

So, $f(x)=-\left(\frac{\pi}{4}-x\right)=x-\frac{\pi}{4}$

Let $y=\frac{x}{2} \Rightarrow f(y)=2 y-\frac{\pi}{4}$

Now, differentiate w.r.t. y, $\frac{d f(y)}{d y}=2$.