Let f be a differentiable function satisfying `f(xy)=f(x).f(y).AA x gt 0, y gt 0` and `f(1+x)=1+x{1+g(x)}`, where `lim_(x to 0)g(x)=0` then `int (f(x))/(f'(x))dx` is equal to

Options

A. `x^2/2 + C`
B. `x^3/2+C`
C. `x^3/3 +C`
D. `x^2/3 + C`

Explanation

To solve the given problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Understand the functional equation
We are given that \( f(xy) = f(x)f(y) \) for all \( x > 0 \) and \( y > 0 \).

**Hint:** This is a property of exponential functions. Consider what kind of functions satisfy this property.

### Step 2: Find \( f(1) \)
Set \( x = 1 \) and \( y = 1 \):
\[ f(1 \cdot 1) = f(1)f(1) \implies f(1) = f(1)^2 \]
This implies \( f(1) = 1 \) (since \( f(1) = 0 \) would not satisfy the functional equation).

**Hint:** Think about the implications of squaring a number and setting it equal to itself.

### Step 3: Differentiate the functional equation
We partially differentiate \( f(xy) = f(x)f(y) \) with respect to \( x \):
\[ \frac{\partial}{\partial x} f(xy) = y f'(xy) = f'(x)f(y) \]
Now, set \( x = 1 \):
\[ y f'(y) = f'(1)f(y) \]
Rearranging gives:
\[ \frac{f(y)}{f'(y)} = \frac{y}{f'(1)} \]

**Hint:** This relationship can be useful for integration later.

### Step 4: Substitute \( y \) with \( x \)
Replacing \( y \) with \( x \):
\[ \frac{f(x)}{f'(x)} = \frac{x}{f'(1)} \]

**Hint:** This is a key relationship that will help us in the next step.

### Step 5: Integrate both sides
Integrate both sides with respect to \( x \):
\[ \int \frac{f(x)}{f'(x)} \, dx = \int \frac{x}{f'(1)} \, dx \]
The right-hand side integrates to:
\[ \frac{1}{f'(1)} \cdot \frac{x^2}{2} + C \]
Thus, we have:
\[ \int \frac{f(x)}{f'(x)} \, dx = \frac{x^2}{2f'(1)} + C \]

**Hint:** Remember to include the constant of integration.

### Step 6: Find \( f'(1) \)
To find \( f'(1) \), we use the second condition \( f(1+x) = (1+x)(1+g(x)) \):
\[ f'(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0} \frac{(1+h)(1+g(h)) - 1}{h} \]
This simplifies to:
\[ = \lim_{h \to 0} \frac{h(1 + g(h))}{h} = 1 + \lim_{h \to 0} g(h) = 1 + 0 = 1 \]

**Hint:** The limit helps us find the derivative at a specific point.

### Step 7: Substitute \( f'(1) \) back into the integral
Now substituting \( f'(1) = 1 \) into our integral result:
\[ \int \frac{f(x)}{f'(x)} \, dx = \frac{x^2}{2 \cdot 1} + C = \frac{x^2}{2} + C \]

**Hint:** This gives us the final form of the integral.

### Conclusion
Thus, the integral \( \int \frac{f(x)}{f'(x)} \, dx \) is equal to:
\[ \frac{x^2}{2} + C \]

### Final Answer
The answer is \( \frac{x^2}{2} + C \).