A function `f : R -> R^+` satisfies `f(x+y)= f(x) f(y) AA x in R` If `f'(0)=2` then `f'(x)=`

Options

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

Answer

To solve the problem step by step, we will use the properties of the function \( f \) given in the question.

### Step 1: Understand the given functional equation
The function \( f \) satisfies the equation:
\[ f(x+y) = f(x) f(y) \quad \forall x, y \in \mathbb{R} \]
This is a well-known property of exponential functions.

**Hint:** This property suggests that \( f \) could be of the form \( f(x) = e^{g(x)} \) for some function \( g \).

### Step 2: Differentiate the functional equation
To differentiate the functional equation, we can set \( y = 0 \):
\[ f(x+0) = f(x) f(0) \]
This simplifies to:
\[ f(x) = f(x) f(0) \]
Assuming \( f(x) \neq 0 \), we can divide both sides by \( f(x) \):
\[ 1 = f(0) \]
Thus, we find that:
\[ f(0) = 1 \]

**Hint:** Setting \( y = 0 \) helps in finding the value of \( f(0) \).

### Step 3: Differentiate both sides with respect to \( y \)
Differentiating the original equation \( f(x+y) = f(x) f(y) \) with respect to \( y \) gives:
\[ \frac{d}{dy} f(x+y) = f'(x+y) \]
On the right-hand side, using the product rule:
\[ \frac{d}{dy} [f(x) f(y)] = f(x) f'(y) \]
Thus, we have:
\[ f'(x+y) = f(x) f'(y) \]

**Hint:** This step uses the chain rule and product rule of differentiation.

### Step 4: Substitute \( y = 0 \) into the differentiated equation
Substituting \( y = 0 \) into the differentiated equation gives:
\[ f'(x+0) = f(x) f'(0) \]
This simplifies to:
\[ f'(x) = f(x) f'(0) \]
Since we know \( f'(0) = 2 \), we can substitute this value:
\[ f'(x) = 2 f(x) \]

**Hint:** This substitution shows how the derivative relates to the function itself.

### Step 5: Solve the differential equation
We have the differential equation:
\[ f'(x) = 2 f(x) \]
This can be solved using separation of variables:
\[ \frac{f'(x)}{f(x)} = 2 \]
Integrating both sides:
\[ \int \frac{f'(x)}{f(x)} \, dx = \int 2 \, dx \]
This gives:
\[ \ln |f(x)| = 2x + C \]
Exponentiating both sides, we find:
\[ f(x) = e^{2x + C} = e^C e^{2x} \]
Let \( k = e^C \), then:
\[ f(x) = k e^{2x} \]

**Hint:** The integration leads us to an exponential function.

### Step 6: Determine the value of \( k \)
Since \( f(0) = 1 \):
\[ f(0) = k e^{2 \cdot 0} = k = 1 \]
Thus, we have:
\[ f(x) = e^{2x} \]

**Hint:** Use the initial condition to find the constant \( k \).

### Step 7: Find \( f'(x) \)
Now we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx} e^{2x} = 2 e^{2x} \]

**Hint:** Differentiating the exponential function gives us the final result.

### Final Answer
Thus, the derivative \( f'(x) \) is:
\[ f'(x) = 2 f(x) \]