if `f(x)=x^x,x in (1,infty)` and `g(x)` be inverse function of `f(x)` then `g^(')(x)` must be equal to

Options

A. `(1)/(x(1+l n (g(x))))`
B. `(1)/(x(1+l nx))`
C. `(1)/(g(x).(1+l n (g(x)))`
D. non existent

Explanation

To find \( g'(x) \) where \( g(x) \) is the inverse function of \( f(x) = x^x \) for \( x \in (1, \infty) \), we can follow these steps:

### Step 1: Understand the relationship between \( f \) and \( g \)

Since \( g(x) \) is the inverse of \( f(x) \), we have:
\[ f(g(x)) = x \]

### Step 2: Differentiate both sides with respect to \( x \)

Using implicit differentiation, we differentiate both sides:
\[ \frac{d}{dx}[f(g(x))] = \frac{d}{dx}[x] \]
By applying the chain rule, we get:
\[ f'(g(x)) \cdot g'(x) = 1 \]

### Step 3: Solve for \( g'(x) \)

From the equation above, we can isolate \( g'(x) \):
\[ g'(x) = \frac{1}{f'(g(x))} \]

### Step 4: Find \( f'(x) \)

We know that:
\[ f(x) = x^x \]
To find \( f'(x) \), we can use logarithmic differentiation:
\[ \ln(f(x)) = x \ln(x) \]
Differentiating both sides:
\[ \frac{f'(x)}{f(x)} = \ln(x) + 1 \]
Thus,
\[ f'(x) = f(x) \cdot (\ln(x) + 1) = x^x \cdot (\ln(x) + 1) \]

### Step 5: Substitute \( g(x) \) into \( f' \)

Now we substitute \( g(x) \) into \( f' \):
\[ f'(g(x)) = g(x)^{g(x)} \cdot (\ln(g(x)) + 1) \]

### Step 6: Write the final expression for \( g'(x) \)

Substituting \( f'(g(x)) \) back into the expression for \( g'(x) \):
\[ g'(x) = \frac{1}{g(x)^{g(x)} \cdot (\ln(g(x)) + 1)} \]

### Final Answer

Thus, the derivative of the inverse function \( g'(x) \) is:
\[ g'(x) = \frac{1}{g(x)^{g(x)} \cdot (\ln(g(x)) + 1)} \]