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

Options

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

Answer

To solve the problem, we need to differentiate the function given by:

\[ y = \tan^{-1}\left(\frac{x}{2}\right) - \cot^{-1}\left(\frac{x}{2}\right) \]

### Step 1: Differentiate \( y \)

Using the differentiation formulas for inverse trigonometric functions:

1. The derivative of \( \tan^{-1}(u) \) is \( \frac{1}{1 + u^2} \cdot \frac{du}{dx} \)
2. The derivative of \( \cot^{-1}(u) \) is \( -\frac{1}{1 + u^2} \cdot \frac{du}{dx} \)

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

Now we can differentiate \( y \):

\[ \frac{dy}{dx} = \frac{d}{dx}\left(\tan^{-1}\left(\frac{x}{2}\right)\right) - \frac{d}{dx}\left(\cot^{-1}\left(\frac{x}{2}\right)\right) \]

Using the chain rule:

\[ \frac{dy}{dx} = \frac{1}{1 + \left(\frac{x}{2}\right)^2} \cdot \frac{1}{2} - \left(-\frac{1}{1 + \left(\frac{x}{2}\right)^2} \cdot \frac{1}{2}\right) \]

### Step 2: Simplify the expression

Now, substituting \( u = \frac{x}{2} \):

\[ \frac{dy}{dx} = \frac{1}{1 + \frac{x^2}{4}} \cdot \frac{1}{2} + \frac{1}{1 + \frac{x^2}{4}} \cdot \frac{1}{2} \]

Combining the two terms:

\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{1 + \frac{x^2}{4}} + \frac{1}{2} \cdot \frac{1}{1 + \frac{x^2}{4}} = \frac{1}{1 + \frac{x^2}{4}} \]

### Step 3: Further simplify

Now, we can simplify the expression:

\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{2}{2 + \frac{x^2}{2}} = \frac{2}{2 + \frac{x^2}{2}} = \frac{4}{4 + x^2} \]

### Final Result

Thus, the derivative \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = \frac{4}{4 + x^2} \]

### Conclusion

The answer is \( \frac{4}{4 + x^2} \).

---