If g is the inverse of f and f'(x) = `(1)/(1 + x^(2))` , then g'(x) is equal to

Options

A. `1 + [g(x)]^(2)`
B. `(-1)/(1 + [g(x)]^(2))`
C. `(1)/(2 (1 + x^(2)))`
D. None of these

Explanation

To find \( g'(x) \) where \( g \) is the inverse of \( f \) and \( f'(x) = \frac{1}{1 + x^2} \), we can use the relationship between the derivatives of inverse functions.

### Step-by-Step Solution:

1. **Understanding the relationship between \( f \) and \( g \)**:
Since \( g \) is the inverse of \( f \), we have:
\[ f(g(x)) = x \]

2. **Differentiate both sides with respect to \( x \)**:
Using the chain rule, we differentiate the left side:
\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]
The right side differentiates to:
\[ \frac{d}{dx}[x] = 1 \]
Therefore, we have:
\[ f'(g(x)) \cdot g'(x) = 1 \]

3. **Solve for \( g'(x) \)**:
Rearranging the equation gives:
\[ g'(x) = \frac{1}{f'(g(x))} \]

4. **Substituting \( f'(x) \)**:
We know that \( f'(x) = \frac{1}{1 + x^2} \). Thus, we substitute \( g(x) \) into \( f' \):
\[ g'(x) = \frac{1}{f'(g(x))} = \frac{1}{\frac{1}{1 + (g(x))^2}} = 1 + (g(x))^2 \]

5. **Final expression for \( g'(x) \)**:
Therefore, we conclude that:
\[ g'(x) = 1 + (g(x))^2 \]