Differentiate `tan^(-1){sqrt((1-cosx)/(1+cosx))},\ -pi ltx lt pi` with respect to `x` :

Explanation

To differentiate the function \( y = \tan^{-1}\left(\sqrt{\frac{1 - \cos x}{1 + \cos x}}\right) \) with respect to \( x \), we will follow these steps:

### Step 1: Simplify the Argument of the Inverse Tangent
We start with the expression inside the inverse tangent:
\[ y = \tan^{-1}\left(\sqrt{\frac{1 - \cos x}{1 + \cos x}}\right) \]
We know that:
\[ 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \quad \text{and} \quad 1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right) \]
Thus, we can rewrite the argument:
\[ \frac{1 - \cos x}{1 + \cos x} = \frac{2 \sin^2\left(\frac{x}{2}\right)}{2 \cos^2\left(\frac{x}{2}\right)} = \tan^2\left(\frac{x}{2}\right) \]
Therefore, we have:
\[ y = \tan^{-1}\left(\sqrt{\tan^2\left(\frac{x}{2}\right)}\right) = \tan^{-1}\left|\tan\left(\frac{x}{2}\right)\right| \]

### Step 2: Determine the Sign of \( \tan\left(\frac{x}{2}\right) \)
In the interval \( -\pi < x < \pi \):
- For \( -\pi < x < 0 \), \( \tan\left(\frac{x}{2}\right) < 0 \)
- For \( 0 < x < \pi \), \( \tan\left(\frac{x}{2}\right) > 0 \)

Thus, we can express \( y \) as:
\[ y = \begin{cases} -\frac{x}{2} & \text{if } -\pi < x < 0 \\ \frac{x}{2} & \text{if } 0 < x < \pi \end{cases} \]

### Step 3: Differentiate the Piecewise Function
Now we differentiate \( y \) for each case:
1. For \( -\pi < x < 0 \):
\[ \frac{dy}{dx} = -\frac{1}{2} \]
2. For \( 0 < x < \pi \):
\[ \frac{dy}{dx} = \frac{1}{2} \]

### Step 4: Check Differentiability at \( x = 0 \)
At \( x = 0 \), we need to check the left-hand limit (LHL) and right-hand limit (RHL):
- LHL:
\[ \lim_{x \to 0^-} \frac{y - 0}{x - 0} = \lim_{x \to 0^-} \frac{-\frac{x}{2}}{x} = -\frac{1}{2} \]
- RHL:
\[ \lim_{x \to 0^+} \frac{y - 0}{x - 0} = \lim_{x \to 0^+} \frac{\frac{x}{2}}{x} = \frac{1}{2} \]

Since LHL \( \neq \) RHL, the function is not differentiable at \( x = 0 \).

### Final Result
Thus, the derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \begin{cases} -\frac{1}{2} & \text{if } -\pi < x < 0 \\ \frac{1}{2} & \text{if } 0 < x < \pi \end{cases} \]
And the function is not differentiable at \( x = 0 \).