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 find \( g'(x) \) given that \( f(x) = x + \tan x \) and \( f \) is the inverse of \( g \), we can follow these steps:

### Step 1: Understand the relationship between \( f \) and \( g \)
Since \( f \) is the inverse of \( g \), we have:
\[ g(f(x)) = x \]
This implies that if we differentiate both sides with respect to \( x \), we can find \( g' \).

### Step 2: Differentiate both sides
Differentiating \( g(f(x)) = x \) using the chain rule gives:
\[ g'(f(x)) \cdot f'(x) = 1 \]

### Step 3: Find \( f'(x) \)
Now, we need to find \( f'(x) \). Since \( f(x) = x + \tan x \), we differentiate:
\[ f'(x) = 1 + \sec^2 x \]

### Step 4: Substitute \( f'(x) \) into the derivative equation
Substituting \( f'(x) \) into the equation from Step 2:
\[ g'(f(x)) \cdot (1 + \sec^2 x) = 1 \]

### Step 5: Solve for \( g'(f(x)) \)
Rearranging gives:
\[ g'(f(x)) = \frac{1}{1 + \sec^2 x} \]

### Step 6: Use the identity \( \sec^2 x = 1 + \tan^2 x \)
Using the trigonometric identity \( \sec^2 x = 1 + \tan^2 x \):
\[ g'(f(x)) = \frac{1}{1 + (1 + \tan^2 x)} = \frac{1}{2 + \tan^2 x} \]

### Step 7: Substitute back \( f(x) \)
Since \( f(x) = x + \tan x \), we can express \( \tan^2 x \) in terms of \( f(x) \):
\[ \tan x = f(x) - x \implies \tan^2 x = (f(x) - x)^2 \]

### Step 8: Final expression for \( g'(x) \)
Thus, we have:
\[ g'(f(x)) = \frac{1}{2 + (f(x) - x)^2} \]

This gives us the required expression for \( g'(x) \).

### Final Answer
\[ g'(x) = \frac{1}{2 + (f(x) - x)^2} \]