If `f(x) = x + tan x` and `f` is the inverse of `g`, then `g'(x)` is equal to

Options

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

Explanation

To solve the problem, we need to find \( g'(x) \) given that \( f(x) = x + \tan x \) and \( f \) is the inverse of \( g \).

### Step-by-Step Solution:

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

2. **Differentiate both sides**:
We differentiate both sides of the equation \( f(g(x)) = x \) with respect to \( x \):
\[ \frac{d}{dx}[f(g(x))] = \frac{d}{dx}[x] \]
Using the chain rule on the left side, we get:
\[ f'(g(x)) \cdot g'(x) = 1 \]

3. **Finding \( f'(x) \)**:
Now, we need to find \( f'(x) \). Given \( f(x) = x + \tan x \), we differentiate:
\[ f'(x) = 1 + \sec^2 x \]

4. **Substituting \( g(x) \) into \( f' \)**:
We substitute \( g(x) \) into \( f' \):
\[ f'(g(x)) = 1 + \sec^2(g(x)) \]

5. **Rearranging the equation for \( g'(x) \)**:
From the equation \( f'(g(x)) \cdot g'(x) = 1 \), we can solve for \( g'(x) \):
\[ g'(x) = \frac{1}{f'(g(x))} = \frac{1}{1 + \sec^2(g(x))} \]

6. **Using the identity \( \sec^2 x = 1 + \tan^2 x \)**:
We can rewrite \( \sec^2(g(x)) \):
\[ g'(x) = \frac{1}{1 + (1 + \tan^2(g(x)))} \]
This simplifies to:
\[ g'(x) = \frac{1}{2 + \tan^2(g(x))} \]

### Final Answer:
Thus, the derivative \( g'(x) \) is given by:
\[ g'(x) = \frac{1}{2 + \tan^2(g(x))} \]