Let f be differentiable function satisfying `f((x)/(y))=f(x) - f(y)"for all" x, y gt 0`. If f'(1) = 1, then f(x) is

Options

A. `2 log_(e) x`
B. `3 log_(e) x`
C. `log_(e) x`
D. `(1)/(2)log_(e)x`

Answer

To solve the problem, we need to find the function \( f(x) \) given the condition \( f\left(\frac{x}{y}\right) = f(x) - f(y) \) for all \( x, y > 0 \) and that \( f'(1) = 1 \).

### Step-by-Step Solution:

1. **Understanding the Functional Equation**:
We start with the functional equation:
\[ f\left(\frac{x}{y}\right) = f(x) - f(y) \]
This resembles the property of logarithmic functions. We will assume a form of \( f(x) \) that might satisfy this equation.

2. **Assuming a Form for \( f(x) \)**:
Let's assume:
\[ f(x) = \alpha \log x \]
for some constant \( \alpha \). We will check if this form satisfies the given functional equation.

3. **Substituting into the Functional Equation**:
Substitute \( f(x) = \alpha \log x \) into the functional equation:
\[ f\left(\frac{x}{y}\right) = \alpha \log\left(\frac{x}{y}\right) = \alpha (\log x - \log y) = \alpha \log x - \alpha \log y = f(x) - f(y) \]
This shows that our assumption holds true for the functional equation.

4. **Finding \( \alpha \) Using the Derivative Condition**:
We know that \( f'(1) = 1 \). First, we compute the derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(\alpha \log x) = \frac{\alpha}{x} \]
Now, substituting \( x = 1 \):
\[ f'(1) = \frac{\alpha}{1} = \alpha \]
Given \( f'(1) = 1 \), we have:
\[ \alpha = 1 \]

5. **Final Form of the Function**:
Substituting \( \alpha = 1 \) back into our assumed function:
\[ f(x) = \log x \]

### Conclusion:
Thus, the function \( f(x) \) is:
\[ \boxed{\log x} \]