If `f (x) = x^(2) +x ^(4) +log x and g ` is the inverse of f, then `g'(2)` is:

Options

A. 8
B. `1/8`
C. 2
D. `1/4`

Answer

To solve the problem, we need to find \( g'(2) \) where \( g \) is the inverse of the function \( f(x) = x^3 + x^4 + \log x \).

### Step-by-Step Solution:

1. **Understand 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:
\[ f'(g(x)) \cdot g'(x) = 1 \]
From this, we can express \( g'(x) \):
\[ g'(x) = \frac{1}{f'(g(x))} \]

3. **Find \( g(2) \)**:
We need to find \( g(2) \). This means we need to find \( x \) such that \( f(x) = 2 \).
\[ f(1) = 1^3 + 1^4 + \log(1) = 1 + 1 + 0 = 2 \]
Therefore, \( g(2) = 1 \).

4. **Substitute \( g(2) \) into the expression for \( g'(x) \)**:
Now we can find \( g'(2) \):
\[ g'(2) = \frac{1}{f'(g(2))} = \frac{1}{f'(1)} \]

5. **Calculate \( f'(x) \)**:
We need to find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(x^3 + x^4 + \log x) = 3x^2 + 4x^3 + \frac{1}{x} \]

6. **Evaluate \( f'(1) \)**:
Now we substitute \( x = 1 \) into \( f'(x) \):
\[ f'(1) = 3(1^2) + 4(1^3) + \frac{1}{1} = 3 + 4 + 1 = 8 \]

7. **Find \( g'(2) \)**:
Now we can find \( g'(2) \):
\[ g'(2) = \frac{1}{f'(1)} = \frac{1}{8} \]

### Final Answer:
Thus, the value of \( g'(2) \) is:
\[ \boxed{\frac{1}{8}} \]