If ` y = tan^(-1) (x)/(2) - cot ^(-1) (x)/(2) , "then" (dy)/(dx) ` is equal to

Options

A. `(2)/(4 + x^(2))`
B. `(8)/(4 + x^(2))`
C. `(4)/(4 + x^(2))`
D. `(1)/(4 + x^(2))`

Explanation

To find the derivative \( \frac{dy}{dx} \) of the function \( y = \tan^{-1}\left(\frac{x}{2}\right) - \cot^{-1}\left(\frac{x}{2}\right) \), we will follow these steps:

### Step 1: Differentiate \( y \)
We start with the given function:
\[ y = \tan^{-1}\left(\frac{x}{2}\right) - \cot^{-1}\left(\frac{x}{2}\right) \]

### Step 2: Use the derivatives of inverse trigonometric functions
The derivatives of the inverse trigonometric functions are:
- \( \frac{d}{dx} \tan^{-1}(u) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \)
- \( \frac{d}{dx} \cot^{-1}(u) = -\frac{1}{1 + u^2} \cdot \frac{du}{dx} \)

Here, let \( u = \frac{x}{2} \). Then, \( \frac{du}{dx} = \frac{1}{2} \).

### Step 3: Differentiate \( \tan^{-1}\left(\frac{x}{2}\right) \)
Using the chain rule:
\[ \frac{d}{dx} \tan^{-1}\left(\frac{x}{2}\right) = \frac{1}{1 + \left(\frac{x}{2}\right)^2} \cdot \frac{1}{2} = \frac{1}{2(1 + \frac{x^2}{4})} = \frac{1}{2 \cdot \frac{4 + x^2}{4}} = \frac{2}{4 + x^2} \]

### Step 4: Differentiate \( -\cot^{-1}\left(\frac{x}{2}\right) \)
Using the chain rule:
\[ \frac{d}{dx} \left(-\cot^{-1}\left(\frac{x}{2}\right)\right) = -\left(-\frac{1}{1 + \left(\frac{x}{2}\right)^2} \cdot \frac{1}{2}\right) = \frac{1}{1 + \frac{x^2}{4}} \cdot \frac{1}{2} = \frac{1}{2(1 + \frac{x^2}{4})} = \frac{2}{4 + x^2} \]

### Step 5: Combine the derivatives
Now, we combine the derivatives:
\[ \frac{dy}{dx} = \frac{2}{4 + x^2} + \frac{2}{4 + x^2} = \frac{4}{4 + x^2} \]

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{4}{4 + x^2} \]