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-by-Step Solution:

1. **Understand the relationship between \( f \) and \( g \)**:
Since \( f \) is the inverse of \( g \), we have:
\[ f(g(x)) = x \]
This means that if we apply \( f \) to \( g(x) \), we should get \( x \).

2. **Express \( f \) in terms of \( g \)**:
Given \( f(x) = x + \tan x \), we can write:
\[ f(g(x)) = g(x) + \tan(g(x)) = x \]

3. **Differentiate both sides with respect to \( x \)**:
Differentiate the equation \( g(x) + \tan(g(x)) = x \):
\[ \frac{d}{dx}[g(x) + \tan(g(x))] = \frac{d}{dx}[x] \]
This gives:
\[ g'(x) + \sec^2(g(x)) \cdot g'(x) = 1 \]

4. **Factor out \( g'(x) \)**:
We can factor \( g'(x) \) out from the left-hand side:
\[ g'(x)(1 + \sec^2(g(x))) = 1 \]

5. **Solve for \( g'(x) \)**:
Now, isolate \( g'(x) \):
\[ g'(x) = \frac{1}{1 + \sec^2(g(x))} \]

6. **Use the identity for secant**:
Recall that \( \sec^2(g(x)) = 1 + \tan^2(g(x)) \). Thus, we can rewrite:
\[ g'(x) = \frac{1}{2 + \tan^2(g(x))} \]

7. **Express \( \tan^2(g(x)) \)**:
From our earlier equation \( g(x) + \tan(g(x)) = x \), we can rearrange to find \( \tan(g(x)) \):
\[ \tan(g(x)) = x - g(x) \]
Therefore,
\[ \tan^2(g(x)) = (x - g(x))^2 \]

8. **Substitute back into \( g'(x) \)**:
Substitute \( \tan^2(g(x)) \) back into the equation for \( g'(x) \):
\[ g'(x) = \frac{1}{2 + (x - g(x))^2} \]

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