Let `f (x) gt 0` for all x and f'(x) exists for all x. If `f` is the inverse functin of h and `h '(x) = (1)/( 1 + log x).` Then `f '(x)` will be

Options

A. `1 + log (f (x))`
B. `1 + f (x)`
C. `1- log (f (x))`
D. `log f (x)`

Answer

To solve the problem, we need to find \( f'(x) \) given that \( f \) is the inverse function of \( h \) and \( h'(x) = \frac{1}{1 + \log x} \).

### Step-by-step Solution:

1. **Understanding the relationship between \( f \) and \( h \)**:
Since \( f \) is the inverse of \( h \), we have:
\[ h(f(x)) = x \]
This means that applying \( h \) to \( f(x) \) gives us back \( x \).

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

3. **Substituting the expression for \( h'(x) \)**:
We know from the problem statement that:
\[ h'(x) = \frac{1}{1 + \log x} \]
Therefore, substituting \( f(x) \) into this expression gives us:
\[ h'(f(x)) = \frac{1}{1 + \log(f(x))} \]

4. **Setting up the equation**:
Now we can substitute \( h'(f(x)) \) back into our differentiated equation:
\[ \frac{1}{1 + \log(f(x))} \cdot f'(x) = 1 \]

5. **Solving for \( f'(x) \)**:
We can isolate \( f'(x) \) by multiplying both sides by \( 1 + \log(f(x)) \):
\[ f'(x) = 1 + \log(f(x)) \]

### Final Result:
Thus, the derivative \( f'(x) \) is given by:
\[ f'(x) = 1 + \log(f(x)) \]