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

Options

A. `(pi)/(4)+(1)/(2)`
B. `(pi)/(4)-(1)/(2)`
C. `(1)/(2)-(pi)/(4)`
D. none of these

Answer

To find \( f'(1) \) for the function \( f(x) = x \tan^{-1}(x) \), we will follow these steps:

### Step 1: Differentiate \( f(x) \)

We will use the product rule for differentiation, which states that if \( f(x) = u(x) v(x) \), then:

\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]

In our case, let \( u(x) = x \) and \( v(x) = \tan^{-1}(x) \).

### Step 2: Find \( u'(x) \) and \( v'(x) \)

1. Differentiate \( u(x) = x \):
\[ u'(x) = 1 \]

2. Differentiate \( v(x) = \tan^{-1}(x) \):
\[ v'(x) = \frac{1}{1 + x^2} \]

### Step 3: Apply the product rule

Now, we can apply the product rule:

\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]

Substituting the derivatives we found:

\[ f'(x) = 1 \cdot \tan^{-1}(x) + x \cdot \frac{1}{1 + x^2} \]

This simplifies to:

\[ f'(x) = \tan^{-1}(x) + \frac{x}{1 + x^2} \]

### Step 4: Evaluate \( f'(1) \)

Now we need to find \( f'(1) \):

\[ f'(1) = \tan^{-1}(1) + \frac{1}{1 + 1^2} \]

Since \( \tan^{-1}(1) = \frac{\pi}{4} \) and \( 1 + 1^2 = 2 \):

\[ f'(1) = \frac{\pi}{4} + \frac{1}{2} \]

### Step 5: Final answer

Thus, the value of \( f'(1) \) is:

\[ f'(1) = \frac{\pi}{4} + \frac{1}{2} \]