Let `f` be a one-one function satisfying `f'(x)=f(x)` then `(f^-1)''(x)` is equal to

Options

A. `-(1)/(x^(3))`
B. `-(1)/(x^(2))`
C. f(x)
D. `f^(-1)(x)`

Explanation

To solve the problem, we need to find the second derivative of the inverse function \( (f^{-1})''(x) \) given that \( f'(x) = f(x) \).

### Step-by-Step Solution:

1. **Understanding the Function**:
We are given that \( f'(x) = f(x) \). This is a differential equation whose solution is known to be of the form:
\[ f(x) = Ce^x \]
where \( C \) is a constant. Since \( f \) is a one-one function, we can take \( C = 1 \) for simplicity, thus:
\[ f(x) = e^x \]

2. **Finding the Inverse Function**:
To find the inverse function \( f^{-1}(x) \), we set:
\[ y = f(x) = e^x \]
To find \( x \) in terms of \( y \), we take the natural logarithm:
\[ x = \ln(y) \]
Therefore, the inverse function is:
\[ f^{-1}(x) = \ln(x) \]

3. **First Derivative of the Inverse Function**:
We now differentiate \( f^{-1}(x) \):
\[ (f^{-1})'(x) = \frac{d}{dx} \ln(x) = \frac{1}{x} \]

4. **Second Derivative of the Inverse Function**:
Next, we differentiate \( (f^{-1})'(x) \) to find the second derivative:
\[ (f^{-1})''(x) = \frac{d}{dx} \left( \frac{1}{x} \right) = -\frac{1}{x^2} \]

5. **Final Result**:
Thus, we find that:
\[ (f^{-1})''(x) = -\frac{1}{x^2} \]

### Conclusion:
The value of \( (f^{-1})''(x) \) is \( -\frac{1}{x^2} \).