The derivative of $$\tan^{-1}\left(\frac{\sin x - \cos x}{\sin x + \cos x}\right)$$ with respect to $$\frac{x}{2}$$, where $$x \in \left(0, \frac{\pi}{2}\right)$$, is

We need to find the derivative of $$\tan^{-1}\left(\frac{\sin x - \cos x}{\sin x + \cos x}\right)$$ with respect to $$\frac{x}{2}$$, where $$x \in \left(0, \frac{\pi}{2}\right)$$.

Let $$y = \tan^{-1}\left(\frac{\sin x - \cos x}{\sin x + \cos x}\right)$$.

We simplify the argument inside the $$\tan^{-1}$$. Divide both numerator and denominator by $$\cos x$$:

$$\frac{\sin x - \cos x}{\sin x + \cos x} = \frac{\tan x - 1}{\tan x + 1}$$

We use the tangent subtraction formula. Recall that $$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$.

Comparing with $$\frac{\tan x - 1}{1 + \tan x \cdot 1}$$, we recognise this as:

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

Therefore:

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

For $$x \in \left(0, \frac{\pi}{2}\right)$$, we have $$x - \frac{\pi}{4} \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$$, which lies within $$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$.

So $$\tan^{-1}(\tan(\theta)) = \theta$$ applies directly, and:

$$y = x - \frac{\pi}{4}$$

Now we find the derivative of $$y$$ with respect to $$\frac{x}{2}$$. Let $$u = \frac{x}{2}$$, so $$x = 2u$$.

Using the chain rule:

$$\frac{dy}{du} = \frac{dy}{dx} \cdot \frac{dx}{du}$$

We have $$\frac{dy}{dx} = 1$$ (since $$y = x - \frac{\pi}{4}$$) and $$\frac{dx}{du} = 2$$ (since $$x = 2u$$).

Therefore:

$$\frac{dy}{d\left(\frac{x}{2}\right)} = 1 \times 2 = 2$$

The correct answer is Option A: 2.