If `f(x)=xtan^(-1)x,` then `f'(1)` is equal to

Options

A. `(1)/(2)+(pi)/(4)`
B. `-(1)/(2)+(pi)/(4)`
C. `-(1)/(2)-(pi)/(4)`
D. `(1)/(2)-(pi)/(4)`

Explanation

To find \( f'(1) \) for the function \( f(x) = x \tan^{-1}(x) \), we will use the product rule of differentiation.

### Step-by-Step Solution:

1. **Identify the Functions**:
We have \( f(x) = x \tan^{-1}(x) \). Here, we can identify \( u = x \) and \( v = \tan^{-1}(x) \).

2. **Differentiate Using the Product Rule**:
The product rule states that if \( f(x) = u \cdot v \), then:
\[ f'(x) = u'v + uv' \]
where \( u' \) is the derivative of \( u \) and \( v' \) is the derivative of \( v \).

3. **Find \( u' \) and \( v' \)**:
- The derivative of \( u = x \) is:
\[ u' = 1 \]
- The derivative of \( v = \tan^{-1}(x) \) is:
\[ v' = \frac{1}{1 + x^2} \]

4. **Apply the Product Rule**:
Now substituting \( u \), \( v \), \( u' \), and \( v' \) into the product rule:
\[ f'(x) = (1)(\tan^{-1}(x)) + (x)\left(\frac{1}{1 + x^2}\right) \]
Simplifying this gives:
\[ f'(x) = \tan^{-1}(x) + \frac{x}{1 + x^2} \]

5. **Evaluate \( f'(1) \)**:
Now we need to find \( f'(1) \):
\[ f'(1) = \tan^{-1}(1) + \frac{1}{1 + 1^2} \]
We know that \( \tan^{-1}(1) = \frac{\pi}{4} \) and \( 1 + 1^2 = 2 \), so:
\[ f'(1) = \frac{\pi}{4} + \frac{1}{2} \]

6. **Final Result**:
Thus, the value of \( f'(1) \) is:
\[ f'(1) = \frac{\pi}{4} + \frac{1}{2} \]